New book: Inequalities with Beautiful Solutions – V. Cirtoaje, V. Q. B. Can, T. Q. Anh

8 12 2009

Because of some inconvenient things, I am now moving all my databases to my new blog: http://can-hang2007.blogspot.com.

You can view this box at this address: http://can-hang2007.blogspot.com/2009/12/new-book-inequalities-with-beautiful_20.html

Very sorry for this change.


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Categories : Inequalities

Inequality 90 [D. Grinberg]

3 12 2009

If a,b,c are positive real numbers, then

\displaystyle \frac{a^{2}(b+c)}{b^{2}+c^{2}}+\frac{b^{2}(c+a)}{c^{2}+a^{2}}+\frac{c^{2}(a+b)}{a^{2}+b^{2}}\geq a+b+c.

 

First proof. We have

\displaystyle \begin{aligned} \sum \left[ \frac{a^{2}(b+c)}{b^{2}+c^{2}}-a\right] &=\sum \frac{ab(a-b)-ca(c-a)}{b^{2}+c^{2}} \\ &=\sum ab(a-b)\left( \frac{1}{b^{2}+c^{2}}-\frac{1}{c^{2}+a^{2}}\right) \\ &=\sum \frac{ab(a+b)(a-b)^{2}}{(a^{2}+c^{2})(b^{2}+c^{2})}\geq 0. \end{aligned}

Thus, it follows that

\displaystyle \sum \left[ \frac{a^{2}(b+c)}{b^{2}+c^{2}}-a\right] \geq 0,

or

\displaystyle \frac{a^{2}(b+c)}{b^{2}+c^{2}}+\frac{b^{2}(c+a)}{c^{2}+a^{2}}+\frac{c^{2}(a+b)}{a^{2}+b^{2}}\geq a+b+c,

which is just the desired inequality. Equality holds if and only if a=b=c.

 

Second proof. Having in view of the identity

\displaystyle \frac{a^{2}(b+c)}{b^{2}+c^{2}}=\frac{(b+c)(a^{2}+b^{2}+c^{2})}{b^{2}+c^{2}}-b-c,

we can write the desired inequality as

\displaystyle \frac{b+c}{b^{2}+c^{2}}+\frac{c+a}{c^{2}+a^{2}}+\frac{a+b}{a^{2}+b^{2}}\geq \frac{3(a+b+c)}{a^{2}+b^{2}+c^{2}}.

Without loss of generality, assume that a\geq b\geq c. Since a^{2}+c^{2}\geq b^{2}+c^{2} and

\displaystyle \frac{b+c}{b^{2}+c^{2}}-\frac{a+c}{a^{2}+c^{2}}=\frac{(a-b)(ab+bc+ca-c^{2})}{(a^{2}+c^{2})(b^{2}+c^{2})}\geq 0,

by Chebyshev’s Inequality, we have

\displaystyle [ (b^{2}+c^{2})+(a^{2}+c^{2})]\left( \frac{b+c}{b^{2}+c^{2}}+\frac{a+c}{a^{2}+c^{2}}\right) \geq 2[(b+c)+(a+c)],

or

\displaystyle \frac{b+c}{b^{2}+c^{2}}+\frac{a+c}{a^{2}+c^{2}}\geq \frac{2(a+b+2c)}{a^{2}+b^{2}+2c^{2}}.

Therefore, it suffices to prove that

\displaystyle \frac{2(a+b+2c)}{a^{2}+b^{2}+2c^{2}}+\frac{a+b}{a^{2}+b^{2}}\geq \frac{3(a+b+c)}{a^{2}+b^{2}+c^{2}},

which is equivalent to the obvious inequality

\displaystyle \frac{c(a^{2}+b^{2}-2c^{2})(a^{2}+b^{2}-ac-bc)}{(a^{2}+b^{2})(a^{2}+b^{2}+c^{2})(a^{2}+b^{2}+2c^{2})}\geq 0.


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Categories : Inequalities

Inequality 89 [V. Q. B. Can, P. H. Duc]

3 12 2009

If a,b,c are positive real numbers, then

\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{9(a^{2}+b^{2}+c^{2})}{(a+b+c)^{2}}.

 

Proof. By applying the known inequality

\displaystyle (x+y+z)^{3}\geq \frac{27}{4}(x^{2}y+y^{2}z+z^{2}x+xyz)\quad\forall x,\text{ }y,\text{ }z\geq 0

for x=\dfrac{a}{b}, y=\dfrac{b}{c} and z=\dfrac{c}{a}, we get

\displaystyle \left( \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) ^{3}\geq \frac{27}{4}\left( \frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}+1\right) =\frac{27}{4}\left( \frac{a^{3}+b^{3}+c^{3}}{abc}+1\right) .

Therefore, it suffices to prove that

\displaystyle \frac{a^{3}+b^{3}+c^{3}}{abc}+1\geq \frac{108(a^{2}+b^{2}+c^{2})^{3}}{(a+b+c)^{6}},

or

\displaystyle \frac{(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)}{abc}+4\geq \frac{108(a^{2}+b^{2}+c^{2})^{3}}{(a+b+c)^{6}}.

Using now the obvious inequality 3abc(a+b+c)\leq (ab+bc+ca)^{2}, we have

\displaystyle \frac{a+b+c}{abc}=\frac{3(a+b+c)^{2}}{3abc(a+b+c)}\geq \frac{3(a+b+c)^{2}}{(ab+bc+ca)^{2}},

and hence, it is enough to check that

\displaystyle \frac{3(a+b+c)^{2}(a^{2}+b^{2}+c^{2}-ab-bc-ca)}{(ab+bc+ca)^{2}}+4\geq \frac{108(a^{2}+b^{2}+c^{2})^{3}}{(a+b+c)^{6}}.

Setting t=\dfrac{3(a^{2}+b^{2}+c^{2})}{(a+b+c)^{2}}, 1\leq t<3. The above inequality is equivalent to

\displaystyle \frac{54(t-1)}{(3-t)^{2}}+4\geq 4t^{3},

or

\displaystyle (t-1)(9-6t-8t^{2}+10t^{3}-2t^{4})\geq 0.

But this is true because

\displaystyle 9-6t-8t^{2}+10t^{3}-2t^{4}=2(3+3t-t^{2})(t-1)^{2}+3>0.

The proof is completed. Equality holds if and only if a=b=c.


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Categories : Inequalities

Inequality 88 [V. Q. B. Can]

3 12 2009

Let a,b,c be positive real numbers. Prove that

\displaystyle \frac{4(a^{2}+b^{2}+c^{2})}{ab+bc+ca}+\sqrt{3}\sum \frac{a+2b}{\sqrt{a^{2}+2b^{2}}}\geq 13.

 

Proof. Denote

\displaystyle P=\frac{a+2b}{\sqrt{a^{2}+2b^{2}}}+\frac{b+2c}{\sqrt{b^{2}+2c^{2}}}+\frac{c+2a}{\sqrt{c^{2}+2a^{2}}}.

By the AM-GM Inequality and the Cauchy-Schwarz Inequality, we have

\displaystyle \begin{aligned}P &=\sqrt{3}\sum \frac{(a+2b)^{2}}{(a+2b)\sqrt{3(a^{2}+2b^{2})}}\geq 2\sqrt{3}\sum \frac{(a+2b)^{2}}{(a+2b)^{2}+3(a^{2}+2b^{2})} \\ &\geq \frac{2\sqrt{3}\left[ \displaystyle\sum (a+2b)\right] ^{2}}{\displaystyle\sum [(a+2b)^{2}+3(a^{2}+2b^{2})]}=\frac{9\sqrt{3}\left( \displaystyle\sum a\right) ^{2}}{\displaystyle7\sum a^{2}+2\sum ab}.\end{aligned}

Therefore, it suffices to prove that

\displaystyle \frac{4(a^{2}+b^{2}+c^{2})}{ab+bc+ca}+\frac{27(a+b+c)^{2}}{7(a^{2}+b^{2}+c^{2})+2(ab+bc+ca)}\geq 13,

which is equivalent to the obvious inequality

\displaystyle \frac{28(a^{2}+b^{2}+c^{2}-ab-bc-ca)^{2}}{(ab+bc+ca)[7(a^{2}+b^{2}+c^{2})+2(ab+bc+ca)]}\geq 0.

Equality holds if and only if a=b=c.


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Categories : Inequalities

Inequality 87 [Unknown author]

3 12 2009

Let a, b, c be positive real numbers such that ab+bc+ca +abc \ge 4. Prove that

\displaystyle \frac{1}{(a+1)^2(b+c)}+\frac{1}{(b+1)^2(c+a)}+\frac{1}{(c+1)^2(a+b)} \le \frac{3}{8}.

 

Proof. Letting a=tx, b=ty and c=tz, where t>0 and x,y,z>0 such that xy+yz+zx+xyz=4. The condition ab+bc+ca+abc \ge 4 implies t \ge 1, and the inequality becomes

\displaystyle \frac{1}{t(tx+1)^2(y+z)}+\frac{1}{t(ty+1)^2(z+x)}+\frac{1}{t(tz+1)^2(x+y)} \le \frac{3}{8}.

We see that it suffices to prove this inequality for t=1. In this case, we may write the inequality in the form

\displaystyle \frac{1}{(x+1)^2(y+z)}+\frac{1}{(y+1)^2(z+x)}+\frac{1}{(z+1)^2(x+y)} \le \frac{3}{8}.

Now, since x, y, z>0 and xy+yz+zx+xyz=4, there exist some positive real numbers u, v, w such that x=\dfrac{2u}{v+w}, y=\dfrac{2v}{w+u} and z=\dfrac{2w}{u+v}. The above inequality becomes

\displaystyle \sum \frac{(u+v)(u+w)(v+w)^2}{(2u+v+w)^2[v(u+v)+w(u+w)]} \le \frac{3}{4}.

Using the AM-GM Inequality and the Cauchy-Schwarz Inequality, we get

\displaystyle \begin{aligned} \frac{(u+v)(u+w)(v+w)^2}{(2u+v+w)^2[v(u+v)+w(u+w)]} &\le \frac{(v+w)^2}{4[v(u+v)+w(u+w)]}\\ & \le \frac{1}{4}\left( \frac{v}{u+v}+\frac{w}{u+w}\right).\end{aligned}

Therefore,

\displaystyle \sum \frac{(u+v)(u+w)(v+w)^2}{(2u+v+w)^2[v(u+v)+w(u+w)]} \le \frac{1}{4}\sum \left( \frac{v}{u+v}+\frac{w}{u+w}\right) =\frac{3}{4}.

The proof is completed. Equality holds if and only if a=b=c=1.

 

Remark. The proof of this problem gives us the fourth proof of the previous problem, because the condition abc \ge 1 implies ab+bc+ca+abc \ge 4.


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Categories : Inequalities

Inequality 86 [T. Q. Anh]

3 12 2009

Let a, b, c be positive real numbers such that abc=1. Prove that

\displaystyle \frac{1}{(a+1)^2(b+c)}+\frac{1}{(b+1)^2(c+a)}+\frac{1}{(c+1)^2(a+b)} \le \frac{3}{8}.

 

First proof. Setting a=x^2,b=y^2 and c=z^2, where x,y,z are positive real numbers. The inequality becomes

\displaystyle \frac{1}{(x^2+1)^2(y^2+z^2)}+\frac{1}{(y^2+1)^2(z^2+x^2)}+\frac{1}{(z^2+1)^2(x^2+y^2)} \le \frac{3}{8}.

By the Cauchy-Schwarz Inequality, we have

\displaystyle \sqrt{(x^2+1)(y^2+z^2)} \ge xy+z=\frac{1}{z}+z=\frac{z^2+1}{z},

and

\displaystyle \sqrt{(1+x^2)(y^2+z^2)} \ge y+xz=y+\frac{1}{y}=\frac{y^2+1}{y}.

Multiplying these two inequalities, we get

\displaystyle (x^2+1)(y^2+z^2) \ge \frac{(y^2+1)(z^2+1)}{yz}.

It follows that

\displaystyle \sum \frac{1}{(x^2+1)^2(y^2+z^2)} \le \sum \frac{yz}{(x^2+1)(y^2+1)(z^2+1)}.

Therefore, it suffices to prove that

\displaystyle (x^2+1)(y^2+1)(z^2+1) \ge \frac{8}{3}(xy+yz+zx).

Using again the Cauchy-Schwarz Inequality, we have

\displaystyle \begin{array}{c}\sqrt{(x^2+1)(1+y^2)} \ge x+y,\quad \sqrt{(y^2+1)(1+z^2)} \ge y+z,\\ \sqrt{(z^2+1)(1+x^2)} \ge z+x.\end{array}

Multiplying these three inequalities and then using the known inequality

\displaystyle (x+y)(y+z)(z+x) \ge \frac{8}{9}(x+y+z)(xy+yz+zx),

we get

\displaystyle \begin{aligned} (x^2+1)(y^2+1)(z^2+1) &\ge (x+y)(y+z)(z+x) \\ &\ge \frac{8}{9}(x+y+z)(xy+yz+zx).\end{aligned}

Therefore, it suffices to show that

x+y+z \ge 3,

which is true according to the AM-GM Inequality. Equality holds if and only if a=b=c=1.

 

Second proof. Setting a=x^3, b=y^3 and c=z^3, we have

\displaystyle \frac{1}{(a+1)^2(b+c)}=\frac{x^3y^3z^3}{(x^3+xyz)^2(y^3+z^3)}=\frac{xy^3z^3}{(x^2+yz)^2(y^3+z^3)}.

By the AM-GM Inequality, we get

\displaystyle (x^2+yz)(y+z)=y(x^2+z^2)+z(x^2+y^2) \ge 2\sqrt{yz(x^2+y^2)(x^2+z^2)}.

This yields

\displaystyle (x^2+yz)^2(y+z) \ge \frac{4yz(x^2+y^2)(x^2+z^2)}{y+z}.

Using this in combination with the obvious inequality 2(y^2-yz+z^2) \ge y^2+z^2, we get

\displaystyle \begin{aligned} \frac{1}{(a+1)^2(b+c)} &\le \frac{xy^2z^2(y+z)}{4(x^2+y^2)(x^2+z^2)(y^2-yz+z^2)} \\ &\le \frac{xy^2z^2(y+z)}{2(x^2+y^2)(x^2+z^2)(y^2+z^2)}.\end{aligned}

Therefore, it suffices to prove that

\displaystyle 4\sum xy^2z^2(y+z) \le 3(x^2+y^2)(y^2+z^2)(z^2+x^2).

Dividing each side of this inequality by x^2y^2z^2, it becomes

\displaystyle 3 \left(\frac{x}{y}+\frac{y}{x}\right)\left(\frac{y}{z}+\frac{z}{y}\right)\left(\frac{z}{x}+\frac{x}{z}\right) \ge 4\sum \left( \frac{x}{y}+\frac{y}{x}\right),

or

\displaystyle 3\left[ 2+\frac{(x-y)^2}{xy}\right]\left[2+\frac{(y-z)^2}{yz}\right]\left[2+\frac{(z-x)^2}{zx}\right] \ge 24+4\sum \frac{(x-y)^2}{xy}.

Now, using the trivial inequality

(2+u)(2+v)(2+w) \ge 8+4(u+v+w) \quad \forall u,v, w \ge 0,

we get

\displaystyle \begin{aligned} 3\left[ 2+\frac{(x-y)^2}{xy}\right]\left[2+\frac{(y-z)^2}{yz}\right]\left[2+\frac{(z-x)^2}{zx}\right] &\ge 24+12\sum \frac{(x-y)^2}{xy}\\ & \ge 24+4\sum \frac{(x-y)^2}{xy}, \end{aligned}

which completes the proof.

 

Third proof. Since a, b, c>0 and abc=1, there exist some positive real numbers x, y, z such that a=\dfrac{y}{x}, b=\dfrac{x}{z} and c=\dfrac{z}{y}. After making this substitution, the inequality becomes

\displaystyle \frac{x^2yz}{(x+y)^2(xy+z^2)}+\frac{y^2zx}{(y+z)^2(yz+x^2)}+\frac{z^2xy}{(z+x)^2(zx+y^2)} \le \frac{3}{8}.

By the AM-GM Inequality, we have

xy+z^2 \ge 2z\sqrt{xy},

and

(x+y)^2 \ge 2\sqrt{2xy(x^2+y^2)}.

Therefore,

\displaystyle \frac{x^2yz}{(x+y)^2(xy+z^2)} \le \frac{x}{4\sqrt{2(x^2+y^2)}}.

It suffices to show that

\displaystyle \frac{x}{\sqrt{x^2+y^2}}+\frac{y}{\sqrt{y^2+z^2}}+\frac{z}{\sqrt{z^2+x^2}} \le \frac{3}{\sqrt{2}},

which is just a known result.

 

Remark. The proofs of this problem gives us various proofs of the previous problem, because we have 4x^x \ge (x+1)^2 for any x>0.


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Categories : Inequalities

Inequality 85 [J. Chen]

27 11 2009

Let a,b,c be positive real numbers such that abc=1. Prove that

\displaystyle \frac{1}{a^{a}(b+c)}+\frac{1}{b^{b}(c+a)}+\frac{1}{c^{c}(a+b)}\leq \frac{3}{2}.

 

Proof. Without loss of generality, assume that c=\min \{a,b,c\}. From abc=1, we get c\leq 1. Thus, by Bernoulli’s Inequality, we have

\displaystyle \frac{1}{c^{c}}=\left( 1+\frac{1}{c}-1\right) ^{c}\leq 1+c\left( \frac{1}{c}-1\right) =2-c.

On the other hand, it is known that x^{x}\geq x for any positive real number x. According these two inequalities, we see that it suffices to show that

\displaystyle \frac{1}{a(b+c)}+\frac{1}{b(a+c)}+\frac{2-c}{a+b}\leq \frac{3}{2}.

Since

\displaystyle \frac{1}{a(b+c)}+\frac{1}{b(a+c)}=\frac{bc}{b+c}+\frac{ac}{a+c}=2c-c^{2}\left( \frac{1}{a+c}+\frac{1}{b+c}\right)

and

\displaystyle \frac{1}{a+c}+\frac{1}{b+c}-\frac{2}{\sqrt{ab}+c}=\frac{\left( \sqrt{a}-\sqrt{b}\right) ^{2}\left( \sqrt{ab}-c\right) }{(a+c)(b+c)\left( \sqrt{ab}+c\right) }\geq 0,

we get

\displaystyle \frac{1}{a(b+c)}+\frac{1}{b(a+c)}\leq 2c-\frac{2c^{2}}{\sqrt{ab}+c}.

Besides, it is clear from the AM-GM Inequality that

\displaystyle \frac{2-c}{a+b}\leq \frac{2-c}{2\sqrt{ab}}.

Therefore, it suffices to prove that

\displaystyle 2c-\frac{2c^{2}}{\sqrt{ab}+c}+\frac{2-c}{2\sqrt{ab}}\leq \frac{3}{2}.

Setting t=\sqrt{ab}, we get c=\dfrac{1}{t^{2}} and the inequality becomes

\displaystyle \frac{2}{t^{2}}-\frac{\dfrac{2}{t^{4}}}{t+\dfrac{1}{t^{2}}}+\frac{2-\dfrac{1}{t^{2}}}{2t}\leq \frac{3}{2},

which is equivalent to the obvious inequality

\displaystyle \frac{(3t^{4}+4t^{3}+t^{2}+2t+1)(t-1)^{2}}{2t^{3}(t^{3}+1)}\geq 0.

The proof is completed. Equality holds if and only if a=b=c=1.


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Categories : Inequalities

Inequality 84 [D. D. Lam]

26 11 2009

Let a,b,c be nonnegative real numbers, no two of which are zero. Prove that

\displaystyle\sum \frac{1}{a^{2}+bc}\geq \sum \frac{1}{a^{2}+2bc}+\frac{ab+bc+ca}{2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})}.

 

Proof. Without loss of generality, assume that a\geq b\geq c. There are two cases to consider: b+c\geq a and a>b+c.

 

Case 1. b+c\geq a. Write the inequality as

\displaystyle\sum \left( \frac{1}{a^{2}+bc}-\frac{1}{a^{2}+2bc}\right) \geq \frac{ab+bc+ca}{2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})},

or

\displaystyle\sum \frac{bc}{(a^{2}+bc)(a^{2}+2bc)}\geq \frac{ab+bc+ca}{2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})}.

By the Cauchy-Schwarz Inequality, we have

\displaystyle\sum \frac{bc}{(a^{2}+bc)(a^{2}+2bc)}\geq \frac{\displaystyle \left( \sum bc\right) ^{2}}{\displaystyle\sum bc(a^{2}+bc)(a^{2}+2bc)},

so it is enough to prove that

\displaystyle2\left( \sum bc\right) \left( \sum b^{2}c^{2}\right) \geq \sum bc(a^{2}+bc)(a^{2}+2bc).

This inequality reduces to

\displaystyle abc\left[ 2\sum ab(a+b)-\sum a^{3}-9abc\right] \geq 0,

or equivalently,

\displaystyle2\sum ab(a+b)\geq \sum a^{3}+9abc.

Setting x=b+c-a\geq 0, y=c+a-b\geq 0 and z=a+b-c\geq 0, we get

\displaystyle a=\frac{y+z}{2},\quad b=\frac{z+x}{2},\quad c=\frac{x+y}{2}.

Substituting into the above inequality, it becomes

x^{3}+y^{3}+z^{3}+3xyz\geq xy(x+y)+yz(y+z)+zx(z+x),

which is the third degree Schur’s Inequality.

 

Case 2. a>b+c\geq 2c. In this case, we write the inequality as

\displaystyle\sum \left[ \frac{1}{a^{2}+bc}-\frac{1}{a^{2}+2bc}-\frac{bc}{2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})}\right] \geq 0,

or

\displaystyle\frac{abc}{2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})}\sum \frac{a(2b^{2}+2c^{2}-3bc-a^{2})}{(a^{2}+bc)(a^{2}+2bc)}\geq 0.

Since

\displaystyle\frac{c(2a^{2}+2b^{2}-3ab-c^{2})}{(c^{2}+ab)(c^{2}+2ab)}\geq 0,

it suffices to show that

\displaystyle\frac{a(2b^{2}+2c^{2}-3bc-a^{2})}{(a^{2}+bc)(a^{2}+2bc)}+\frac{b(2c^{2}+2a^{2}-3ca-b^{2})}{(b^{2}+ca)(b^{2}+2ca)}\geq 0.

We have

\begin{aligned}\displaystyle\frac{b}{(b^{2}+ca)(b^{2}+2ca)}&-\frac{a}{(a^{2}+bc)(a^{2}+2bc)}=\\ &\displaystyle =\frac{(a^{3}-b^{3})(ab-2c^{2})}{(a^{2}+bc)(b^{2}+ca)(a^{2}+2bc)(b^{2}+2ca)}\geq 0 \end{aligned}

and

2c^{2}+2a^{2}-3ca-b^{2}=(a^{2}-b^{2})+(a-c)(a-2c)\geq 0,

hence we get

\displaystyle\frac{b(2c^{2}+2a^{2}-3ca-b^{2})}{(b^{2}+ca)(b^{2}+2ca)}\geq \frac{a(2c^{2}+2a^{2}-3ca-b^{2})}{(a^{2}+bc)(a^{2}+2bc)}.

Thus, it is enough to check that

(2b^{2}+2c^{2}-3bc-a^{2})+(2c^{2}+2a^{2}-3ca-b^{2})\geq 0,

or

a^{2}+b^{2}+4c^{2}-3c(a+b)\geq 0.

This inequality is true because

a^{2}+b^{2}+4c^{2}-3c(a+b)=(a+b-2c)(a-b-c)+2(b-c)^{2}\geq 0.

The proof is completed. Equality holds if and only if a=b=c, or a=0, or b=0, or c=0.


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Categories : Inequalities

Inequality 83 [Unknown author]

26 11 2009

If \displaystyle a,b,c,d are nonnegative real numbers, then

\displaystyle \sum a^{4}+\sum abc(a+b+c)\geq 2\sum_{sym}a^{2}b^{2}+4abcd.

 

Proof. Without loss of generality, assume that \displaystyle d=\min \{a,b,c,d\}. Using the fourth degree Schur’s Inequality, we have

\displaystyle \begin{aligned} \sum_{a,b,c}a^{4}+abc(a+b+c) &\geq \sum_{a,b,c}ab(a^{2}+b^{2}) \\ \displaystyle &=2\sum_{a,b,c}a^{2}b^{2}+\sum_{a,b,c}ab(a-b)^{2} \\ \displaystyle &\geq 2\sum_{a,b,c}a^{2}b^{2}+d^{2}\sum_{a,b,c}(a-b)^{2}.\end{aligned}

Therefore, it suffices to prove that

\displaystyle d^{2}\sum_{a,b,c}(a-b)^{2}+d^{4}+d\sum_{a,b,c}ab(a+b)+d^{2}\sum_{a,b,c}ab\geq 2d^{2}\sum_{a,b,c}a^{2}+4abcd,

or

\displaystyle d^{3}+\sum_{a,b,c}ab(a+b)\geq d(ab+bc+ca)+4abc.

By the AM-GM Inequality, we have

\displaystyle \sum_{a,b,c}ab(a+b)\geq 6abc,

so it is enough to check that

\displaystyle d^{3}+2abc-d(ab+bc+ca)\geq 0,

which is equivalent to the obvious inequality

\displaystyle d(b-d)(c-d)+c(a-d)(b-d)+b(a-d)(c-d)\geq 0.

The proof is completed. On the assumption \displaystyle d=\min \{a,b,c,d\}, equality holds for \displaystyle a=b=c=d, and again for \displaystyle a=b=c and \displaystyle d=0.


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Categories : Inequalities

Inequality 82 [Unknown author]

26 11 2009

If a,b,c,d are nonnegative real numbers, then

(a+b+c+d)^{3}\geq 4[a(c+d)^{2}+b(d+a)^{2}+c(a+b)^{2}+d(b+c)^{2}].

 

Proof. Setting

P(a,b,c,d)=a(c+d)^{2}+b(d+a)^{2}+c(a+b)^{2}+d(b+c)^{2}.

We will show that in order to prove the original inequality, it suffices to prove it for (a-c)(b-d)\geq 0. Indeed, if (a-c)(b-d)<0, then we can set a_{1}=b,b_{1}=c,c_{1}=d,d_{1}=a to have

(a_{1}-c_{1})(b_{1}-d_{1})=-(a-c)(b-d)>0,

and

P(a_{1},b_{1},c_{1},d_{1})=P(a,b,c,d).

Now, using the AM-GM Inequality, we get

\begin{aligned} (a+b+c+d)^{3} &\geq 4(a+d)(b+c)(a+b+c+d) \\ &=4(a+d)^{2}(b+c)+4(b+c)^{2}(a+d).\end{aligned}

Therefore, it suffices to prove that

c(a+d)^{2}+a(b+c)^{2}\geq c(a+b)^{2}+a(c+d)^{2}.

This inequality reduces to

(b+d)(a-c)(b-d)\geq 0,

which is obviously true, and the proof is completed. Equality holds if and only if a=c and b=d.


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Categories : Inequalities

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