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Let $a,b,c>0$ be real numbers such that $a+b+c=3$,how to prove that? :

$$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$$

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Also, why do you think it is true? – Thomas Andrews Jul 9 '12 at 18:07
    
By contradiction ,we can substitute each term in the LHS by x,y,z and we want to prove that $x+y+z \ge1$ but for contradiction we assume $x+y+z<1$,so i think its right – Frank Jul 9 '12 at 19:58
    
With this reasoning, every statement is false? – Did Jul 10 '12 at 6:14
    
No,we want to prove something but we assume the opposite to prove what we want by contradiction . – Frank Jul 10 '12 at 11:28
1  
This is not an indication that you tried anything at all. – Did Jul 12 '12 at 19:23
up vote 5 down vote accepted

By AM>GM $$ \frac{a+b+c}{3}=1 \Rightarrow abc\le 1 \\ \Rightarrow \frac{1}{2ab^2+1} = \frac{1}{2abc\frac{b}{c}+1}\ge\frac{1}{2\frac{b}{c}+1}=\frac{c}{2b+c} \\ S = \frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge\frac{c}{2b+c}+\frac{a}{2c+a}+\frac{b}{2a+b} $$ $$ \begin{align} 3S-3 & \ge \frac{3c-2b-c}{2b+c}+\frac{3a-2c-a}{2c+a}+\frac{3b-2a-b}{2a+b}\\ & = 2\left(\frac{c-b}{2b+c}+\frac{a-c}{2c+a}+\frac{b-a}{2a+b}\right) \\ & = \frac{2}{D}\left(3ab^2+3bc^2+3ca^2-9abc\right)\\ & = \frac{6abc}{D}\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}-3\right) \\ & \ge 0 \end{align} $$ where for clarity we simply write $D$ for the positive denominator, and the last inequality is again by AM>GM $$ 1=\left(\frac{b}{c}\cdot\frac{c}{a}\cdot\frac{a}{b}\right)^{1/3}\le\frac{1}{3}\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}\right) $$ Finally $3S-3\ge 0 \Rightarrow S\ge 1$.

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A solution for straightforward mathematicians. :-)

It seems the following.

After multiplication of both sides of the inequality by a common denominator and simplification, we reduce the initial inequality to

$$1+ab^2+bc^2+ca^2\ge 4a^3b^3c^3.$$

Since $$ab^2+bc^2+ca^2\ge 3abc$$ it suffices to check that

$$1+3abc-4(abc)^3\ge 0,$$

that is

$$(1-abc)(2abc+1)^2\ge 0.$$

Tthe last inequality holds because $$abc\le\left(\frac{a+b+c}3\right)^3=1.$$

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