Prove that for all positive real numbers $a,b,c$ we have:


So far I have solved for $a=b=c=1$ and $a=1$, $b=2$, $c=3$ to show that the inequality holds true but I am needing the answer written more like a proof, I'm just not sure what theorems to apply actually prove the inequality holds true.

  • $\begingroup$ Is the $v$ in the denominator of the fourth term supposed to be a $c$? $\endgroup$
    – Tom
    Aug 5, 2016 at 18:12
  • $\begingroup$ yeah, I made a mistake when mathjaxing it. $\endgroup$
    – Asinomás
    Aug 5, 2016 at 18:13

3 Answers 3


For all real numbers it's wrong, of course.

For positives $a$, $b$ and $c$, we can use an Integral method:

We see that for all positives $x$, $y$ and $z$ the following inequality holds. $$\sum\limits_{cyc}(x^4-2x^2y^2+x^2yz)\geq0$$ Indeed, by Schur $$\sum\limits_{cyc}(x^4-2x^2y^2+x^2yz)=\sum\limits_{cyc}(x^4-x^3y-x^3z+x^2yz)+\sum\limits_{cyc}xy(x-y)^2\geq0$$ Hence, $\sum\limits_{cyc}\left(t^{4a}-2t^{2a+2b}+t^{2a+b+c}\right)\geq0$, where $t>0$ or


Thus, $\int\limits_{0}^1\sum\limits_{cyc}\left(t^{4a-1}-2t^{2a+2b-1}+t^{2a+b+c-1}\right)dt\geq0$,

which gives your inequality. Done!


Also we can use a full expanding: $$\sum\limits_{sym}(2a^6b^2-2a^6bc+9a^5b^3+7a^4b^4-3a^5b^2c+13a^3b^3c-9a^4b^2c^2-17a^3b^3c^2)\geq0$$

which is obviously true by Muirhead.


Also we can use $uvw$ here.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, we need to prove that $\frac{3v^2}{4w^3}+\frac{\sum\limits_{cyc}(3u+a)(3u+b)}{\prod\limits_{cyc}(3u+a)}\geq\frac{\sum\limits_{cyc}(a+b)(a+c)}{9uv^2-w^3}$ or

$\frac{v^2}{4w^3}+\frac{15u^2+v^2}{54u^3+9uv^2+w^3}\geq\frac{3u^2-v^2}{9uv^2-w^3}$, which is $f(w^3)\geq0$,

where $f$ is a concave function (it's obvious that the coefficient before $w^6$ is negative).

But a concave function gets a minimal value for an extremal value of $w^3$,

which happens in the following cases.

  1. $w^3\rightarrow0^+$. In this case our inequality is obviously true.

  2. $b=c=1$, which gives $(a-1)^2\geq0$. Done!


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