Feb 12

quyhktn-qa1's inequality

Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$.Prove that
$\dfrac{1}{13a^2+(b-c)^2}+\dfrac{1}{13b^2+(c-a)^2}+\dfrac{1}{13c^2+(a-b)^2} \geq \dfrac{3}{13}.$

Proof:(by scaleye,taiwan)

The inequality is equivalent to $\frac{1}{13a^2+(b-c)^2}+\frac{1}{13b^2+(c-a)^2}+\frac{1}{13c^2+(a-b)^2} \geq \frac{27}{13(a+b+c)^2}.$
By Cauchy Schwarz Inequality,
\begin{align*} \sum\frac{1}{13a^2+(b-c)^2}&=\sum\frac{(a+13b+13c)^2}{\left[13a^2+(b-c)^2\right](a+13b+13c)^2}\\ &\ge \frac{27^2(a+b+c)^2}{\sum\left[13a^2+(b-c)^2\right](a+13b+13c)^2}. \end{align*}
It suffices to show that
$13\cdot27(a+b+c)^4 \ge \sum\left[13a^2+(b-c)^2\right](a+13b+13c)^2.$
After expanding and simplifying, it is equivalent to
$65\sum(a^3b+ab^3) \ge 122\sum a^2b^2 + 8\sum a^2bc$
or
$61\sum ab(a-b)^2 +2\sum(a^3b+ab^3 +a^3c +ac^3 -4a^2bc) \ge 0.$
Which is true by AM-GM.
Equality holds when $a=b=c=1$  or $a=3$, $b=c=0$ and any  permutations

Feb 12

Unusual inequality, x+y+z=0

Given real numbers $x,y,z$ such that $x+y+z=0$, show that
$\frac{x(x+2)}{2x^2+1}+\frac{y(y+2)}{2y^2+1}+\frac{z(z+2)}{2z^2+1}\ge 0$

proof:(Vo Quoc Ba Can)

Notice that
$\frac{x(x+2)}{2x^2+1}=\frac{(2x+1)^2}{2(2x^2+1)}-\frac{1}{2}.$
So the inequality can be written as
$\sum \frac{(2x+1)^2}{2x^2+1} \ge 3.$
Now, using the Cauchy-Schwarz inequality, we see that
$2x^2=\frac{4}{3}x^2+\frac{2}{3}(y+z)^2 \le \frac{4}{3}x^2+\frac{4}{3}(y^2+z^2).$
Therefore,
$\sum \frac{(2x+1)^2}{2x^2+1} \ge 3\sum \frac{(2x+1)^2}{4(x^2+y^2+z^2)+3}=3.$

Jan 30

an interesting inequality

tian27546 proposed a nice inequality and I give a proof.

Problem

Proof
(1)by Holder inequality,
$\left(\sum{\frac{x^4}{y^3}}\right)^{5}\left(\sum{x^{10}y^5}\right)\left(\sum{x^5y^5}\right)^{2}\geq (\sum{x^5})^{8}$
So,it's suffice to prove
$(\sum{x^5})^8\geq 3^{5}\left(\sum{x^{10}y^5}\right)\left(\sum{x^5y^5}\right)^{2}$
We have the know result
$(\sum{a})^2\geq 3\sum{ab}$
and
$(\sum{a})^5\geq 27(ab+bc+ca)(a^2b+b^2c+c^2a)$
The result follows.
Done!

(2)
After homogenous,just need to check
$\left(\sum{\frac{x^3}{y^2}}\right)^5\geq 81(x^5+y^5+z^5)$
recalling the Well-known Vasile Cirtoaje inequality
$\boxed{(a+b+c)^5\geq 81abc(a^2+b^2+c^2)}$
and use $a=\frac{x^3}{y^2},b=\frac{y^3}{z^2},c=\frac{y^3}{x^2}$,it's suffices to prove
$\frac{x^6}{y^4}+\frac{y^6}{z^4}+\frac{z^6}{x^4}\geq \frac{x^4}{yz}+\frac{y^4}{xz}+\frac{z^4}{xy}$
Now,Using AM-GM inequality
$26\frac{x^6}{y^4}+11\frac{y^6}{z^4}+\frac{z^6}{x^4}\geq 38\frac{x^4}{yz}$
$26\frac{y^6}{z^4}+11\frac{z^6}{x^4}+\frac{x^6}{y^4}\geq 38\frac{y^4}{xz}$
$26\frac{z^6}{x^4}+11\frac{x^6}{y^4}+\frac{y^6}{z^4}\geq 38\frac{z^4}{xy}$
sum its up ,the result follows.
Done!

after a while.he also show the general one

Jan 30

some nice inequality books

1.Vo Quoc Ba Can's book

2.Algebraic Inequality- Old and New method

Author: Vasile Cirtoaje

Jan 30

A New Note On Cauchy-Schwarz

Here is my article about pqr lemma and Cauchy-Schwarz.