 # Prove that ((2a+b+c)^2)/(2a^2+(b+c)^2)+((2b+c+a)^2)/(2b^2+(c+a)^2)+((2c+a+b))/(2c^2+(a+b)^2)<=8 Tamara Bryan 2022-07-21 Answered
Prove that $\frac{\left(2a+b+c{\right)}^{2}}{2{a}^{2}+\left(b+c{\right)}^{2}}+\frac{\left(2b+c+a{\right)}^{2}}{2{b}^{2}+\left(c+a{\right)}^{2}}+\frac{\left(2c+a+b{\right)}^{2}}{2{c}^{2}+\left(a+b{\right)}^{2}}\le 8$
Prove that
$\frac{\left(2a+b+c{\right)}^{2}}{2{a}^{2}+\left(b+c{\right)}^{2}}+\frac{\left(2b+c+a{\right)}^{2}}{2{b}^{2}+\left(c+a{\right)}^{2}}+\frac{\left(2c+a+b\right)}{2{c}^{2}+\left(a+b{\right)}^{2}}\le 8$
MY ATTEMPT:I want to make a relation between $a,b,c$. By trial I found that if we put $a=b=c=1$ then the above inequality holds(equality also holds). So by trial I assume that $a+b+c=3$. After that the three functions become of the form of the function below: $f\left(x\right)=\frac{\left(x+3{\right)}^{2}}{2{x}^{2}+\left(3-x{\right)}^{2}}$. I calculate the function and found that : $f\left(x\right)\le ⅓\left(4x+4\right)$. Am I do right . Anybody has other ideas.
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For non-negatives $a$, $b$ and $c$ let $a+b+c=3$. Hence,
$8-\sum _{cyc}\frac{\left(2a+b+c{\right)}^{2}}{2{a}^{2}+\left(b+c{\right)}^{2}}=\sum _{cyc}\left(\frac{8}{3}-\frac{\left(a+3{\right)}^{2}}{2{a}^{2}+\left(3-a{\right)}^{2}}\right)=$
$=\frac{1}{3}\sum _{cyc}\frac{\left(a-1\right)\left(7a-15\right)}{{a}^{2}-2a+3}=\frac{1}{3}\left(\sum _{cyc}\frac{\left(a-1\right)\left(7a-15\right)}{{a}^{2}-2a+3}+4\left(a-1\right)\right)=$
$=\sum _{cyc}\frac{\left(a-1{\right)}^{2}\left(4a+3\right)}{3\left({a}^{2}-2a+3\right)}\ge 0$
Also there is the following.
It's enough to prove that
$\frac{\left(2a+b+c{\right)}^{2}}{2{a}^{2}+\left(b+c{\right)}^{2}}\le \frac{4\left(4a+b+c\right)}{3\left(a+b+c\right)},$
which is $\left(2a-b-c{\right)}^{2}\left(5a+b+c\right)\ge 0$

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