2
$\begingroup$

Does anyone know how to show that for any $a,b,c>0$,

$$ (a^3+b^3+c^3+9abc)(a+b+c) \geq 4(ab+ac+bc)^2 $$

I could only think of Newton’s inequality, which unfortunately goes the other way : it shows that $abc\ \times \frac{a+b+c}{3} \leq \left( \frac{ab+ac+bc}{3}\right)^2$.

$\endgroup$
2
$\begingroup$

By third degree Schur's inequality, we have $$a^3+b^3+c^3+3abc \ge \sum_{cyc} ab(a+b)$$ So it is enough to show that $$\left(\sum_{cyc} ab(a+b)+6abc\right)(a+b+c) \ge 4(ab+bc+ca)^2$$ which is equivalent on expansion (ugh) to the obvious $$\sum_{cyc}ab(a-b)^2 \ge 0$$


P.S. Equality will be like in Schur, i.e. when $a=b=c$ (or when two of the variables are equal and the other zero, if that's allowed).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.