Let's demonstrate your love with Inequalities

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

Vo Quoc Ba Can — Tue, 08 Dec 2009 03:09:00 +0000

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.

Inequality 90 [D. Grinberg]

Vo Quoc Ba Can — Thu, 03 Dec 2009 22:54:47 +0000

If are positive real numbers, then

First proof. We have

Thus, it follows that

which is just the desired inequality. Equality holds if and only if

Second proof. Having in view of the identity

we can write the desired inequality as

Without loss of generality, assume that Since and

by Chebyshev’s Inequality, we have

Therefore, it suffices to prove that

which is equivalent to the obvious inequality

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

Vo Quoc Ba Can — Thu, 03 Dec 2009 13:21:06 +0000

If are positive real numbers, then

Proof. By applying the known inequality

for and we get

Therefore, it suffices to prove that

Using now the obvious inequality we have

and hence, it is enough to check that

Setting The above inequality is equivalent to

But this is true because

0.' title='\displaystyle 9-6t-8t^{2}+10t^{3}-2t^{4}=2(3+3t-t^{2})(t-1)^{2}+3>0.' class='latex' />

The proof is completed. Equality holds if and only if

Inequality 88 [V. Q. B. Can]

Vo Quoc Ba Can — Thu, 03 Dec 2009 11:33:21 +0000

Let be positive real numbers. Prove that

Proof. Denote

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

Therefore, it suffices to prove that

which is equivalent to the obvious inequality

Equality holds if and only if

Inequality 87 [Unknown author]

Vo Quoc Ba Can — Thu, 03 Dec 2009 11:10:45 +0000

Let be positive real numbers such that Prove that

Proof. Letting and where 0' title='t>0' class='latex' /> and 0' title='x,y,z>0' class='latex' /> such that The condition implies and the inequality becomes

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

Now, since 0' title='x, y, z>0' class='latex' /> and there exist some positive real numbers such that and The above inequality becomes

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

Therefore,

The proof is completed. Equality holds if and only if

Remark. The proof of this problem gives us the fourth proof of the previous problem, because the condition implies

Inequality 86 [T. Q. Anh]

Vo Quoc Ba Can — Thu, 03 Dec 2009 10:54:21 +0000

Let be positive real numbers such that Prove that

First proof. Setting and where are positive real numbers. The inequality becomes

By the Cauchy-Schwarz Inequality, we have

and

Multiplying these two inequalities, we get

It follows that

Therefore, it suffices to prove that

Using again the Cauchy-Schwarz Inequality, we have

Multiplying these three inequalities and then using the known inequality

we get

Therefore, it suffices to show that

which is true according to the AM-GM Inequality. Equality holds if and only if

Second proof. Setting and we have

By the AM-GM Inequality, we get

This yields

Using this in combination with the obvious inequality we get

Therefore, it suffices to prove that

Dividing each side of this inequality by it becomes

Now, using the trivial inequality

we get

which completes the proof.

Third proof. Since 0' title='a, b, c>0' class='latex' /> and there exist some positive real numbers such that and After making this substitution, the inequality becomes

By the AM-GM Inequality, we have

and

Therefore,

It suffices to show that

which is just a known result.

Remark. The proofs of this problem gives us various proofs of the previous problem, because we have for any 0.' title='x>0.' class='latex' />

Inequality 85 [J. Chen]

Vo Quoc Ba Can — Fri, 27 Nov 2009 22:09:55 +0000

Let be positive real numbers such that Prove that

Proof. Without loss of generality, assume that From we get Thus, by Bernoulli’s Inequality, we have

On the other hand, it is known that for any positive real number According these two inequalities, we see that it suffices to show that

Since

and

we get

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

Therefore, it suffices to prove that

Setting we get and the inequality becomes

which is equivalent to the obvious inequality

The proof is completed. Equality holds if and only if

Inequality 84 [D. D. Lam]

Vo Quoc Ba Can — Thu, 26 Nov 2009 19:20:10 +0000

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

Proof. Without loss of generality, assume that There are two cases to consider: and b+c.' title='a>b+c.' class='latex' />

Case 1. Write the inequality as

By the Cauchy-Schwarz Inequality, we have

so it is enough to prove that

This inequality reduces to

or equivalently,

Setting and we get

Substituting into the above inequality, it becomes

which is the third degree Schur’s Inequality.

Case 2. b+c\geq 2c.' title='a>b+c\geq 2c.' class='latex' /> In this case, we write the inequality as

Since

it suffices to show that

We have

and

hence we get

Thus, it is enough to check that

This inequality is true because

The proof is completed. Equality holds if and only if or or or

Inequality 83 [Unknown author]

Vo Quoc Ba Can — Thu, 26 Nov 2009 08:55:00 +0000

If are nonnegative real numbers, then

Proof. Without loss of generality, assume that Using the fourth degree Schur’s Inequality, we have

Therefore, it suffices to prove that

By the AM-GM Inequality, we have

so it is enough to check that

which is equivalent to the obvious inequality

The proof is completed. On the assumption equality holds for and again for and

Inequality 82 [Unknown author]

Vo Quoc Ba Can — Thu, 26 Nov 2009 07:47:33 +0000

If are nonnegative real numbers, then

Proof. Setting

We will show that in order to prove the original inequality, it suffices to prove it for Indeed, if then we can set to have

0,' title='(a_{1}-c_{1})(b_{1}-d_{1})=-(a-c)(b-d)>0,' class='latex' />

and

Now, using the AM-GM Inequality, we get

Therefore, it suffices to prove that

This inequality reduces to

which is obviously true, and the proof is completed. Equality holds if and only if and