Prove that $\frac{bc}{a^2+bc}+\frac{ca}{b^2+ca}+\frac{ab}{c^2+ab}\le{\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}}$

Question

Let $a,b,c$ be positive real numbers. Prove that

$\frac{bc}{a^2+bc}+\frac{ca}{b^2+ca}+\frac{ab}{c^2+ab}\le{\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}}$.

I have no idea how to solve this question.

Answer 1

We can use SOS here. $\sum\limits_{cyc}\left(\frac{a}{b+c}-\frac{bc}{a^2+bc}\right)=\sum\limits_{cyc}\frac{a^3+abc-b^2c-bc^2}{(b+c)(a^2+bc)}=\sum\limits_{cyc}\frac{a^3-abc+2abc-b^2c-bc^2}{(b+c)(a^2+bc)}=$ $=\frac{1}{2}\sum\limits_{cyc}\frac{2a(a^2-bc)+2bc(2a-b-c)}{(b+c)(a^2+bc)}=\frac{1}{2}\sum\limits_{cyc}\frac{(a-b)(a^2+ac+2bc)-(c-a)(a^2+ab+2bc)}{(b+c)(a^2+bc)}=$ $=\frac{1}{2}\sum\limits_{cyc}(a-b)\left(\frac{a^2+ac+2bc}{(b+c)(a^2+bc)}-\frac{b^2+bc+2ac}{(a+c)(b^2+ac)}\right)=\frac{1}{2}\sum\limits_{cyc}\frac{(a-b)^2(a^2b^2+(a^3-a^2b-ab^2+b^3)c+abc^2+ac^3+bc^3)}{(a+c)(b+c)(a^2+bc)(b^2+ac)}\geq0$

Answer 2

This proof appears to be a bit complex.

$ \sum _{ cyc }^{ { } }f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b)$

Note that the $(LHS)\le \sum _{ cyc }^{ { } } \frac { (\frac { b+c }{ 2 } )^{ 2 } }{ a^{ 2 }+(\frac { b+c }{ 2 } )^{ 2 } } (\because AM-GM)$

WLOG $a+b+c=3$

Since $\sum _{ cyc }^{ }{ \frac { 3-3a }{ 2 } } =0$, we can conclude that we need to show $\frac { (\frac { b+c }{ 2 } )^{ 2 } }{ a^{ 2 }+(\frac { b+c }{ 2 } )^{ 2 } } \le \frac { a }{ b+c } +\frac { 3-3a }{ 2 }= \frac { a }{ 3-a } +\frac { 3-3a }{ 2 } $

Note that $\frac { (\frac { b+c }{ 2 } )^{ 2 } }{ a^{ 2 }+(\frac { b+c }{ 2 } )^{ 2 } } =\frac { { (3-a) }^{ 2 } }{ 4a^{ 2 }+(3-a)^{ 2 } } \le \frac { a }{ 3-a } +\frac { 3-3a }{ 2 } \Leftrightarrow \frac { { (3-a) }^{ 2 } }{ 4a^{ 2 }+(3-a)^{ 2 } } \le \frac { 2a+(a-3)(3a-3) }{ 2(3-a) }$

This indicates $2(3-a)^{ 3 }\le (3a^{ 2 }-10a+9)(5a^{ 2 }-6a+9)\Leftrightarrow 15a^{ 4 }-68a^{ 3 }+132a^{ 2 }-144a+81+2(a-3)^{ 3 }\ge 0\\ \Leftrightarrow 15a^{ 4 }-66a^{ 3 }+114a^{ 2 }-90a+27\ge 0\Leftrightarrow 3(a-1)^{ 2 }(5a^{ 2 }-12a+9)\ge 0$

However, since $5a^{ 2 }-12a+9\ge 0$, our inequality is proven.

Answer 3

$$\sum \left(\frac{a}{b+c} - \frac{ab}{c^2+ab}\right) = (a^3+b^3+c^3+abc) \sum {\frac {ca \left( {b}^{2}+{c}^{2} \right) \left( a-b \right) ^{2}}{\displaystyle \prod (a+b)(a^2+bc)}} \geqslant 0.$$