# Prove $\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\leq{3\over2}(a+b+c)$ with $a,b,c$ are nonnegative

Hope someone can help on this inequality using nonanalytical method (i.e. simple elementary method leveraging basic inequalities are prefered).

Prove $\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\leq{3\over2}(a+b+c)$ with $a,b,c$ are nonnegative.

Here is what I already got.

First of all, one should notice equality holds not when $a=b=c$ as one might initially thought. Rather, the equality holds when $(a,b,c)$ is a permutation of $({1\over2},{1\over2},0)$.

Secondly and obviously, this is cyclical and homgeneous and hence we can apply the EMV theorem developed by a IMO golden medalist (reference here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1130901) and then the original inequality can be easily proved by assuming $$f(a,b,c)={\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\over {a+b+c}},$$and then prove $$f(1,1,1)\leq{3\over2},\\f(a,b,0)\leq{3\over2},\forall a,b\geq0.$$

But this kind of proof does not fit my appetite as it not only involves some additional theorem but also not very nice as the simple nice form of the question of itself.

So can people help on some elementary proof that might only taking use of the basic inequalities like AM-GM, Jensen's inequality, etc.

first we check $abc=0$,if $a=0 ,\implies 2\sqrt{bc}\le b+c$,it is true.the "=" will hold when $b=c$

in case $abc \not=0$

let $c=Max${$a,b,c$},

we discuss two cases:

1. $a^2+b^2+c^2 \ge 2(ab+bc+ac) \implies c\ge a+b+2\sqrt{ab} \implies ab \le \dfrac{c^2}{16}$

LHS $\le\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+\dfrac{c^2}{16}}\iff \sqrt{a^2+bc}+\sqrt{b^2+ac}+\dfrac{\sqrt{17}c}{4} \le \dfrac{3(a+b+c)}{2} \iff\sqrt{a^2+bc}+\sqrt{b^2+ac} \le \dfrac{3(a+b)+(3-\dfrac{\sqrt{17}}{2})c}{2} \iff 2 \sqrt{a^2+bc}\sqrt{b^2+ac}\le p^2c^2+q(a+b)c+5a^2+5b^2+18ab \\ p=3-\dfrac{\sqrt{17}}{2}>0.8,q=14-3\sqrt{17}>1\\ \iff 4a^2b^2+4b^3c+4a^3c+4abc^2 \le \left( p^2c^2+q(a+b)c+5a^2+5b^2+18ab\right)^2$

look at RHS, from$18ab$ we take $2ab$ to cover $4a^2b^2$, $16ab+p^2c^2 \implies 32p^2abc^2 >4abc^2,2*5a^2*qac >4a^3c,2*5b^2*qbc>4b^3c \implies LHS <RHS$

2 . $a^2+b^2+c^2 \le 2(ab+bc+ac)$ (that is why we exculde $abc=0$)

with $\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3} \le \sqrt{\dfrac{x+y+z}{3}}$

LHS $\le \sqrt{3(a^2+bc+b^2+ac+c^2+ab)} \iff \\ \sqrt{3(a^2+b^2+c^2+ab+bc+ac)} \le \dfrac{3(a+b+c)}{2} \iff a^2+b^2+c^2 \le 2(ab+bc+ac)$

but first "=" will be hold when$a=b=c$, second "=" will hold when $c=a+b+2\sqrt{ab}$

so it is impossible to hold "=".

QED

By C-S $$\left(\sum_{cyc}\sqrt{a^2+bc}\right)^2\leq\sum_{cyc}\frac{a^2+bc}{2a^2+b^2+c^2+bc}\sum_{cyc}(2a^2+b^2+c^2+bc).$$ Thus, it remains to prove that $$\sum_{cyc}\frac{a^2+bc}{2a^2+b^2+c^2+bc}\sum_{cyc}(4a^2+bc)\leq\frac{9(a+b+c)^2}{4}$$ or $$\sum_{cyc}(2a^8+2a^7b+2a^5c+9a^6b^2+9a^6c^2-23a^5b^3-23a^5c^3+20a^4b^4+11a^6bc+$$ $$+70a^5b^2c+70a^5c^2b+79a^4b^3c+79a^4c^3b+263a^4b^2c^2-195a^3b^3c^2)\geq0,$$ which is obvious.

Done!

• Why this is obvious? Assume a=c=1, then expression = $20b^4 -159b^3+...$ may not be always greater than zero? – Wangwang Oct 1 '17 at 3:05

Proof.

(i) if abc=0, W.L.O.G assume c=0, $\Longleftrightarrow a+b \geq 2\sqrt{ab}$, it is clearly true.

(ii) in the following, assume $abc > 0$.

Case 1: $a^2+b^2+c^2 \leq 2(ab+bc+ca)$

according to Cauchy-Schwarz inequality, we have: $$LHS \leq \sqrt{3(a^2+b^2+c^2+ab+bc+ca)}$$ $$\Longleftrightarrow \sqrt{3(a^2+b^2+c^2+ab+bc+ca)} \leq {3\over2}(a+b+c)$$ $$\Longleftrightarrow 3(a^2+b^2+c^2+ab+bc+ca) \leq {9\over4}(a+b+c)^2$$ $$\Longleftrightarrow a^2+b^2+c^2 \leq 2(ab+bc+ca)$$

it is clearly true.

Case 2: $a^2+b^2+c^2 \geq 2(ab+bc+ca)$

W.L.O.G. assume c=max{a,b,c}, according to Cauchy-Schwarz inequality, $$\sqrt{a^2+bc} + \sqrt{b^2+ac} \leq \sqrt{2(a^2+b^2+ac+bc)}$$

and $$\sqrt{c^2+ab}=\sqrt{c^2+ab}-c+c=\frac{ab}{\sqrt{c^2+ab}+c} +c \leq \frac{ab}{2c} +c$$

$LHS \leq \sqrt{2(a^2+b^2+ac+bc)} + \frac{ab}{2c} +c \leq {3\over2}(a+b+c)$

$$\Longleftrightarrow \sqrt{2(a^2+b^2+ac+bc)} \leq {3\over2}(a+b+c) - c - \frac{ab}{2c}$$ $$\Longleftrightarrow \sqrt{2(a^2+b^2+ac+bc)} \leq {3\over2}(a+b) + {1\over2}c - \frac{ab}{2c}$$ $$\Longleftrightarrow 2(a^2+b^2+ac+bc) \leq ({3\over2}(a+b) + {1\over2}c - \frac{ab}{2c})^2$$ $$\Longleftrightarrow \frac{a^2b^2-6a^2bc-6ab^2c+a^2c^2+16abc^2+b^2c^2-2ac^3-2bc^3+c^4}{4c^2} \geq 0$$ $$\Longleftrightarrow \frac{a^2b^2+abc(16c-6a-6b)+c^2(a^2+b^2+c^2-2ac-2bc)}{4c^2} \geq 0$$

it is also true.