# Non-Symmetric Algebraic inequality

Let $a,b,c$ be positive real numbers. Prove that : $$\frac{ab}{3a+b}+\frac{bc}{2c+b}+\frac{ac}{c+2a}\le \frac{2a+20b+27c}{49}$$

• In order to get the best possible answers, it is helpful to state what your thoughts and attempts on the problem are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – projectilemotion Jun 30 '17 at 16:57

Let $a=6x$, $b=3y$ and $c=2z$.
Hence, we need to prove that $$\frac{xy}{6x+y}+\frac{yz}{3y+4z}+\frac{zx}{z+6x}\leq\frac{2x+10y+9z}{49},$$ which is true by AM-GM.
Indeed, $$\frac{xy}{6x+y}+\frac{yz}{3y+4z}+\frac{zx}{z+6x}\leq\frac{x+6y}{49}+\frac{4y+3z}{49}+\frac{6z+x}{49}=\frac{2x+10y+9z}{49}.$$ Done!