# Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$

Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$

It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq ab+bc+ca+3\sqrt[3]{a^2b^2c^2}$. Then do I use the rearrangement inequality similarly on $3\sqrt[3]{a^2b^2c^2}$?

• Sadly enough, by AM-GM, $ab + bc + ca \geq 3\sqrt[3]{a^2b^2c^2}$ so it's gonna be a bit harder than that. Jan 5, 2016 at 22:43

Yet another way in which this can be shown using Schur's inequality in tandem with the AM-GM inequality is as follows: $$a^2+b^2+c^2+3(a^2b^2c^2)^{1/3}\geqslant a^{2/3}b^{4/3} + a^{4/3}b^{2/3} +b^{2/3}c^{4/3} + b^{4/3}c^{2/3} + a^{2/3}c^{4/3} + a^{4/3}c^{2/3} \\[2ex]= 2\left({a^{2/3}b^{4/3} + a^{4/3}b^{2/3}\over 2} + {b^{2/3}c^{4/3} + b^{4/3}c^{2/3}\over 2} + {a^{2/3}c^{4/3} + a^{4/3}c^{2/3}\over 2}\right)\\[2ex] \geqslant 2(ab + bc + ac)$$

Let $a=x^3$, $b=y^3$, $c=z^3$, then it can be rewritten as: $$x^6+y^6+z^6+3 x^2 y^2 z^2-2 \left(x^3 y^3+x^3 z^3+y^3 z^3\right)\geq 0$$ Use the following notations: $$S_{3}:=xyz\qquad S_2:=xy+yz+xz\qquad S_1=x+y+z$$ Then: $$x^6+y^6+z^6=S_1^6-6 S_2 S_1^4+6 S_3 S_1^3+9 S_2^2 S_1^2-12 S_2 S_3 S_1-2 S_2^3+3 S_3^2$$ $$x^3 y^3+x^3 z^3+y^3 z^3=S_2^3-3 S_1 S_3 S_2+3 S_3^2$$ $$3x^2y^2z^2=3S_3^2$$ Then we only have to prove: $$S_1^6-6 S_2 S_1^4+6 S_3 S_1^3+9 S_2^2 S_1^2-6 S_2 S_3 S_1-4 S_2^3\geq 0$$ Now put $S_2=S_1^2$, and notice that with this: $$\left.S_1^6-6 S_2 S_1^4+6 S_3 S_1^3+9 S_2^2 S_1^2-6 S_2 S_3 S_1-4 S_2^3\right|_{S_2=S_1^2}=0$$ Thus this can be factorised as: $$\left(S_1^2-S_2\right) \left(S_1^4-5 S_2 S_1^2+6 S_3 S_1+4 S_2^2\right)\geq0$$ Since: $(x+y+z)^2\geq 3(xy+yz+xz)\Rightarrow S_1^2\geq 3S_2$ by rearrangement, it is enough to prove that the second factor is non-negative. Return to our previous notations, enough to show: $$x^4+y^4+z^4+(x+y+z)xyz-x^3y-y^3x-y^3z-z^3y-x^3z-xz^3=$$ $$=x^2(x-y)(x-z)+y^2(y-x)(y-z)+z^2(z-x)(z-y)\geq 0$$ Which is trivially true by applying Schur's inequality

$x^3=a^2,y^3=b^2,z^3=c^2 \implies x^3+y^3+z^2 +3xyz \ge 2(\sqrt{(xy)^3}+\sqrt{(yz)^3}+\sqrt{(xz)^3})$

we have $x^3+y^3+z^3 +3xyz \ge xy(x+y)+yz(y+z)+xz(x+z)$

$xy(x+y)\ge 2xy\sqrt{xy}=2\sqrt{(xy)^3} \implies xy(x+y)+yz(y+z)+xz(x+z)\ge 2(\sqrt{(xy)^3}+\sqrt{(yz)^3}+\sqrt{(xz)^3})$

• Uumm...and how is this different from the solution I posted?
– ki3i
Jan 6, 2016 at 2:20

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

Hence, our inequality it's $f(v^2)\geq0$, where $f(v^2)=3u^2-4v^2+w^2$.

Thus, $f$ is a linear function, which says that $f$ get's a minimal value for an extremal value of $v^2$, which happens for equality case of two variables.

Let $b=a=x^3$ and $c=1$.

Hence, we need to prove that $x^6+2+3x^2\geq2(2x^3+1)$,

which is $(x-1)^2(x^2+2x+3)x^2\geq0$. Done!

I think I have got close to proving this but I'm not sure if this is a valid proof - but here goes...

Using AM-GM we can show that: $$a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\tag{1}$$ $$ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\tag{2}$$ We can also show that: $$a^2+b^2+c^2\ge ab+bc+ca\tag{3}$$ We can therefore infer that: $$a^2+b^2+c^2\ge ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\tag{4}$$ Using these facts we can say that: $$a^2+b^2+c^2=3\sqrt[3]{a^2b^2c^2}+\delta_1\text{ where }\delta_1\ge0\tag{5}$$ $$ab+bc+ca=3\sqrt[3]{a^2b^2c^2}+\delta_2\text{ where }\delta_2\ge0\tag{6}$$ We can then use (4) to further infer that: $$\delta_1\ge\delta_2\ge0$$ Finaly we can state that: \begin{align} a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2}&=6\sqrt[3]{a^2b^2c^2}+\delta_1\\ &\ge6\sqrt[3]{a^2b^2c^2}+\delta_2\\ &\ge2\left(3\sqrt[3]{a^2b^2c^2}+\frac{\delta_2}{2}\right)\\ &\ge2\left(ab+bc+ca-\frac{\delta_2}{2}\right)\\ \end{align} I am hoping this aproach triggers a thought in someones brain who can then come up with the final proof.

• Wouldn't the equality case cause the deltas to vanish? That suffices, doesn't it? Jan 6, 2016 at 2:03