proof based on cauchy schwartz inequality

Question

If a,b,c are positive real numbers,prove that $$ \frac{a}{b+2c} + \frac{b}{c+2a} + \frac{c}{a+2b} \ge 1 $$ I tried solving and i have no idea how to proceed I mechanically simplified it it looks promising but im still stuck. This is from the excersice on Cauchy Schwartz Inequality.

Answer 1

Using Cauchy-Schwarz in Engel form we get $$\frac{a^2}{ab+2ac}+\frac{b^2}{bc+ab}+\frac{c^2}{ac+2ac}\geq \frac{(a+b+c)^2}{3ab+3ac+3bc}\geq 1$$ if $$(a+b+c)^2\geq 3ab+3ac+3bc$$ and this is $$a^2+b^2+c^2\geq ab+bc+ca$$

Answer 2

As we have, $2+a^3=a^3+1+1\geqslant 3a$, $b^2+1\geqslant 2b$, thus$$\dfrac{a}{a^3+b^2+c}=\frac{a}{3+a^3+b^2-a-b}\leqslant\frac{a}{3a+2b-a-b}=\frac{a}{2a+b}.$$Similarly, we can get $$\dfrac{b}{b^3+c^2+a}\leqslant\frac{b}{2b+c},\,\,\,\,\,\dfrac{c}{c^3+a^2+b}\leqslant\frac{c}{2c+a}.$$It suffices to show$$\frac{a}{2a+b}+\frac{b}{2b+c}+\frac{c}{2c+a}\leqslant 1\iff \frac{b}{2a+b}+\frac{c}{2b+c}+\frac{a}{2c+a}\geqslant 1$$ By Cauchy's inequality, we get$$(b(2a+b)+c(2b+c)+x(2c+a))\left(\frac{b}{2a+b}+\frac{c}{2b+c}+\frac{a}{2c+a}\right)\geqslant (a+b+c)^2.$$ As $b(2a+y)+z(2b+c)+a(2c+a)=(a+b+c)^2$,

So,

$$\frac{b}{2a+b}+\frac{c}{2b+c}+\frac{a}{2c+a}\geqslant 1$$