Prove that $a^2+b^2+c^2 \geq ab(a+b+\sqrt{ab})+cb(c+b+\sqrt{cb})+ ac(a+c+\sqrt{ac} )$

Question

the question

Let $a,b,c$ be positive numbers such that $a+b+c=1$. Prove that $a^2+b^2+c^2 \geq ab(a+b+\sqrt{ab})+cb(c+b+\sqrt{cb})+ ac(a+c+\sqrt{ac} )$.

the idea

After i put some values to $a$, $b$, and $c$ I got to the conclusion that the case of equality only happens if $a=b=c=\frac{1}{3}$, which got me to the idea that I should use the inequality of means:

$$\min \leq \frac{2ab}{a+b} \leq \sqrt{ab} \leq \frac{a+b}{2} \leq \sqrt{\frac{a^2+b^2}{2}} \leq \max$$

I came to the conclusion that we can demonstrate that:

$\frac{a^2+b^2}{2} \geq ab(a+b+\sqrt{ab})$

we already know by the inequality of means that

$\frac{a^2+b^2}{2} \geq ab*1= ab*(a+b+c)$

The problem would be solved if we show that $a+b+c \geq a+b+\sqrt{ab}$, which I don't think I can demonstrate

We can show an inequality maybe using again the inequality of means....idk

Hope one of you can help me! Thank you!

Answer 1

Multiply LHS by $a+b+c=1$. You get after expansion:

$$\sum a^3 + \sum(a^2b+ab^2)\ge \sum (a^2b+ab^2) +\sum a^{3/2}b^{3/2},$$

which is true due to Muirhead or just $x^2+y^2+z^2\ge xy+yz+zx.$

Answer 2

Using cyclic (not symmetric) summations: $$\begin{align} \sum a^2 &= (\sum a)(\sum a^2) \\ &= \sum a^3 + \sum a^2b+\sum ab^2 \\ &\ge \sum (ab)^{3/2} + \sum a^2b+\sum ab^2 \\ & = \sum ab(a+b+\sqrt{ab}) \end{align}$$ where the inequality is true by AM-GM on $(a^3+b^3)/2$