I can't prove it

leedt26

Given $a,b,c\ge0.$ Prove that:
$a^2+b^2+c^2+\frac{\sqrt{3}.\sqrt[3]{abc}(ab+bc+ca)}{\sqrt{a^2+b^2+c^2}}\ge 2(ab+bc+ca)$

**Posted:** Thu Nov 05, 2009 2:32 am

can_hang2007

**Posted:** Thu Nov 05, 2009 6:26 am

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leedt26

**Posted:** Thu Nov 05, 2009 6:48 am

nguoivn

The first proof of can_hang is nicer than my proof (the second) Very Happy

**Posted:** Thu Nov 05, 2009 7:04 am

Tourish · Hodge Conjecture Offline Joined: 11 Sep 2009 Posts: 53

I think it's quite easy.We can suppoes $ab+bc+ca=1$ ,then it becomes
$a^2+b^2+c^2+\frac{\sqrt{3}\cdot\sqrt[3]{abc}}{\sqrt{a^2+b^2+c^2}}\geq 2$
(i) if $a^2+b^2+c^2\geq 2$ ,it is obvious.
(ii) if $1\leq a^2+b^2+c^2\leq 2$ ,then $\sqrt{3}\leq a+b+c\leq 2$ .
by Schur we have $r\geq\frac{4p-p^3}{9}$ ,then it's enough to prove that
$p^2-2+\frac{\sqrt{3}\cdot\sqrt[3]{\frac{4p-p^3}{9}}}{\sqrt{p^2-2}}\geq 2$
$\Longleftrightarrow (4-p^2)(p^2-2)^{\frac{3}{2}}[(4-p^2)^2(p^2-2)^{\frac{3}{2}}-\frac{\sqrt{3}}{3}p]\leq0$
$\Longleftrightarrow (4-p^2)^4(p^2-2)^3\leq \frac{1}{3}p^2$
By AM-GM we have
$(4-p^2)^4(p^2-2)^3\leq (4-p^2)\cdot\left(\frac{3(4-p^2)+3(p^2-2)}{6}\right)^6=4-p^2\leq \frac{1}{3}p^2$
which is true because $3\leq p^2\leq 4$ ,where $p=a+b+c,q=ab+bc+ca,r=abc$ .

Sad

I always can't open the links given by Can.....

**Posted:** Fri Nov 06, 2009 7:58 am

new_member

Hey tourish,since you can't open the link i just decided to copy it here so you can also see the solution Smile

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

$\begin{aligned} \displaystyle \frac{\sqrt{3}\sqrt[3]{abc}(ab+bc+ca)}{\sqrt{a^{2}+b^{2}+c^{2}}} &\geq \frac{3\sqrt{3}abc}{...$

Therefore, it suffices to prove that

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

Using some simple computations, we can write this inequality as

$\displaystyle \sum a^{4}+abc\sum a\geq \sum ab(a^{2}+b^{2})$ ,

which is true because it is the fourth degree Schur’s Inequality. Equality holds if and only if $a=b=c, or a=b and c=0$ , or any cyclic permutation.

**Posted:** Fri Nov 06, 2009 8:09 am

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can_hang2007

**Posted:** Fri Nov 06, 2009 9:03 am

_________________
The love makes us stronger!

V. Q. B. Can

Tourish · Hodge Conjecture Offline Joined: 11 Sep 2009 Posts: 53

Thank you so much,new_member and Can Smile

**Posted:** Sat Nov 07, 2009 10:37 pm