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Let a,b,c be positive real number, such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that :

$\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2} \leq \frac{3}{16}$

Can anyone help me how to deal with it?

Ewin
  • 1,764

2 Answers2

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since $$(x+y)^2\ge 4xy$$ so $$(a+b+2c)^2=(a+c+b+c)^2\ge 4(a+c)(b+c)$$ so $$\sum_{cyc}\dfrac{16}{(a+b+2c)^2}\le\sum_{cyc}\dfrac{4}{(a+c)(b+c)}=\dfrac{8(a+b+c)}{(a+b)(b+c)(a+c)}$$ and use this well know inequality $$(a+b)(b+c)(c+a)\ge\dfrac{8}{9}(a+b+c)(ab+bc+ac)$$ so $$\dfrac{8(a+b+c)}{(a+b)(b+c)(a+c)}\le\dfrac{9}{ab+bc+ac}\cdots (1)$$ since $$a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Longrightarrow 3(ab+bc+ac)=3abc(a+b+c)\le (ab+bc+ac)^2$$ so $$ab+bc+ac\ge 3$$ so $$\sum_{cyc}\dfrac{16}{(a+b+2c)^2}\le\dfrac{9}{ab+bc+ac}\le 3$$

math110
  • 93,304
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Since $$\displaystyle \frac{1}{a} + \frac{1}{b} + \frac{1}{c}=a+b+c$$ $$\Rightarrow \frac{ab+bc+ca}{abc \left( a+b+c \right)} = 1 $$ $$\Rightarrow (ab+bc+ca)^2=3abc(a+b+c)$$ $$\Rightarrow \frac{3}{ab+bc+ca} = \frac{ab+bc+ca}{abc \left( a+b+c \right)} = 1$$ Or $$\frac{9}{16 \left( ab+bc+ca \right)} = \frac{3}{16}$$ Need proof $$\dfrac{1}{\left( 2a+b+c \right)^2}+\dfrac{1}{\left( 2b+c+a \right)^2}+ \dfrac{1}{\left( 2c+a+b \right)^2} \le \frac{9}{16 \left( ab+bc+ca \right)} $$ Use Am-Gm have $$\left( 2a+b+c \right)^2 = \left( a+b + a+c \right)^2 \ge 4 \left( a+b \right) \left( a+c \right)$$ So $$\dfrac{1}{\left( 2a+b+c \right)^2}+\dfrac{1}{\left( 2b+c+a \right)^2}+ \dfrac{1}{\left( 2c+a+b \right)^2} \le \frac{a+b+c}{ 2 \left( a+b \right) \left( b+c \right) \left( c+a \right)} \le \frac{9}{16 \left( ab+bc+ca \right)}$$ $$ \Leftrightarrow a^2b+ab^2+b^2c+bc^2+c^2a+ca^2 \ge 6abc$$. This is true follow to inequality AM-GM