Tamara Bryan

2022-07-21

Prove that $\frac{(2a+b+c{)}^{2}}{2{a}^{2}+(b+c{)}^{2}}+\frac{(2b+c+a{)}^{2}}{2{b}^{2}+(c+a{)}^{2}}+\frac{(2c+a+b{)}^{2}}{2{c}^{2}+(a+b{)}^{2}}\le 8$

Prove that

$\frac{(2a+b+c{)}^{2}}{2{a}^{2}+(b+c{)}^{2}}+\frac{(2b+c+a{)}^{2}}{2{b}^{2}+(c+a{)}^{2}}+\frac{(2c+a+b)}{2{c}^{2}+(a+b{)}^{2}}\le 8$

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: $f(x)=\frac{(x+3{)}^{2}}{2{x}^{2}+(3-x{)}^{2}}$. I calculate the function and found that : $f(x)\le \u2153(4x+4)$. Am I do right . Anybody has other ideas.

Prove that

$\frac{(2a+b+c{)}^{2}}{2{a}^{2}+(b+c{)}^{2}}+\frac{(2b+c+a{)}^{2}}{2{b}^{2}+(c+a{)}^{2}}+\frac{(2c+a+b)}{2{c}^{2}+(a+b{)}^{2}}\le 8$

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: $f(x)=\frac{(x+3{)}^{2}}{2{x}^{2}+(3-x{)}^{2}}$. I calculate the function and found that : $f(x)\le \u2153(4x+4)$. Am I do right . Anybody has other ideas.

kamphundg4

Beginner2022-07-22Added 20 answers

For non-negatives $a$, $b$ and $c$ let $a+b+c=3$. Hence,

$8-\sum _{cyc}\frac{(2a+b+c{)}^{2}}{2{a}^{2}+(b+c{)}^{2}}=\sum _{cyc}(\frac{8}{3}-\frac{(a+3{)}^{2}}{2{a}^{2}+(3-a{)}^{2}})=$

$=\frac{1}{3}\sum _{cyc}\frac{(a-1)(7a-15)}{{a}^{2}-2a+3}=\frac{1}{3}(\sum _{cyc}\frac{(a-1)(7a-15)}{{a}^{2}-2a+3}+4(a-1))=$

$=\sum _{cyc}\frac{(a-1{)}^{2}(4a+3)}{3({a}^{2}-2a+3)}\ge 0$

Also there is the following.

It's enough to prove that

$\frac{(2a+b+c{)}^{2}}{2{a}^{2}+(b+c{)}^{2}}\le \frac{4(4a+b+c)}{3(a+b+c)},$

which is $(2a-b-c{)}^{2}(5a+b+c)\ge 0$

$8-\sum _{cyc}\frac{(2a+b+c{)}^{2}}{2{a}^{2}+(b+c{)}^{2}}=\sum _{cyc}(\frac{8}{3}-\frac{(a+3{)}^{2}}{2{a}^{2}+(3-a{)}^{2}})=$

$=\frac{1}{3}\sum _{cyc}\frac{(a-1)(7a-15)}{{a}^{2}-2a+3}=\frac{1}{3}(\sum _{cyc}\frac{(a-1)(7a-15)}{{a}^{2}-2a+3}+4(a-1))=$

$=\sum _{cyc}\frac{(a-1{)}^{2}(4a+3)}{3({a}^{2}-2a+3)}\ge 0$

Also there is the following.

It's enough to prove that

$\frac{(2a+b+c{)}^{2}}{2{a}^{2}+(b+c{)}^{2}}\le \frac{4(4a+b+c)}{3(a+b+c)},$

which is $(2a-b-c{)}^{2}(5a+b+c)\ge 0$

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