Show that $\displaystyle abc\le (ab+bc+ca)(a^2+b^2+c^2)^2$ for all positive real $a,\,b$ and $c$ such that $a+b+c=1.$

Show that $\displaystyle abc\le (ab+bc+ca)(a^2+b^2+c^2)^2$ for all positive real $a,\,b$ and $c$ such that $a+b+c=1.$

My solution:

$\displaystyle (ab+bc+ca)(a^2+b^2+c^2)^2$

$\displaystyle \ge \frac{(ab+bc+ca)(a^2+b^2+c^2)(a+b+c)^2}{3}$ since $3(a^2+b^2+c^2)\ge (a+b+c)^2$

$\displaystyle = \frac{(ab+bc+ca)(a^2+b^2+c^2)}{3}$ since $a+b+c=1$

$\displaystyle = \frac{a^3b+b^3c+ac^3+a^3c+ab^3+bc^3+a^2bc+ab^2c+abc^2}{3}$

$\displaystyle = \frac{\left(\frac{a^2}{\frac{1}{ab}}+\frac{b^2}{\frac{1}{bc}}+\frac{c^2}{\frac{1}{ca}}\right)+\left(\frac{a^2}{\frac{1}{ac}}+\frac{b^2}{\frac{1}{ab}}+\frac{c^2}{\frac{1}{bc}}\right)+a^2bc+ab^2c+abc^2}{3}$

$\displaystyle \ge \frac{\left(\frac{(a+b+c)^2}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\right)+\left(\frac{(a+b+c)^2}{\frac{1}{ac}+\frac{1}{ab}+\frac{1}{bc}}\right)+a^2bc+ab^2c+abc^2}{3}$ (By the Titu's Lemma)

$\displaystyle = \frac{\left(\frac{1}{\frac{a+b+c}{abc}}\right)+\left(\frac{1}{\frac{a+b+c}{abc}}\right)+a^2bc+ab^2c+abc^2}{3}$

(since $\displaystyle a+b+c=1$ and $\displaystyle \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{c}{abc}+\frac{a}{abc}+\frac{b}{cab}=\frac{a+b+c}{abc})$

$\displaystyle = \frac{\left(abc+abc\right)+abc(a+b+c)}{3}$

$\displaystyle = \frac{3abc}{3}$

$\displaystyle = abc$ (Q.E.D)