## Sunday, June 26, 2016

### Prove that [MATH]\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}\le \frac{1}{abc}[/MATH] for all positive real $a,\,b$ and $c$.

Prove that [MATH]\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}\le \frac{1}{abc}[/MATH] for all positive real $a,\,b$ and $c$.

My solution:

[MATH]\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}[/MATH]

[MATH]=\frac{1}{(a+b)(a^2-ab+b^2)+abc}+\frac{1}{(b+c)(b^2-bc+c^2)+abc}+\frac{1}{(c+a)(c^2-ca+a^2)+abc}[/MATH]

[MATH]\le \frac{1}{(a+b)(2ab-ab)+abc}+\frac{1}{(b+c)(2bc-bc)+abc}+\frac{1}{(c+a)(2ca-ca)+abc}[/MATH] (from AM-GM inequality)

[MATH]= \frac{1}{(a+b)ab+abc}+\frac{1}{(b+c)bc+abc}+\frac{1}{(c+a)ca+abc}[/MATH]

[MATH]= \frac{1}{ab(a+b+c)}+\frac{1}{bc(a+b+c)}+\frac{1}{ca(a+b+c)}[/MATH]

[MATH]= \frac{1}{a+b+c}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)[/MATH]

[MATH]= \frac{1}{a+b+c}\left(\frac{c}{abc}+\frac{a}{abc}+\frac{b}{cab}\right)[/MATH]

[MATH]= \frac{1}{abc}\left(\frac{1}{a+b+c}\right)\left(a+b+c\right)[/MATH]

[MATH]= \frac{1}{abc}[/MATH] (Q.E.D.)

Equality occurs when $a=b=c.$