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How to prove this inequality?

If $a^{2}+b^{2}+c^{2}\leq 3$ and $a,b,c\in \Bbb R^+$, then

$$\left( a+b+c\right) \left( a+b+c-abc\right)\geq 2\left( a^{2}b+b^{2}c+c^{2}a\right) $$

I tried AM>GM but I couldn't get result

Ethan Bolker
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FMath
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    Related http://math.stackexchange.com/questions/270610/inequality-a2b2c2-leq-a2bb2cc2a1 –  Jun 14 '16 at 13:36
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    @Bacon Not really related, at least not any more than any other three variable inequality on this site. For this inequality there is equality for $a = b = c = 1$, while for the one you linked, equality is achieved for $1,0,0$. That would, at least on the surface of things, drastically change the approach to the problem. – Arthur Jun 14 '16 at 14:03

3 Answers3

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Assume w.l.o.g that $a \leq b \leq c$. Then we have : $(a+b+c)^3 \geq 7(a^2b+b^2c+c^2a)+6abc$; thus it is sufficient to prove: $(a+b+c)(a+b+c-abc) \geq \dfrac{2}{7}((a+b+c)^3-6abc)$.

Let $x=a+b+c$ and $y=abc$. The last inequality is then: $x^2-xy \geq \dfrac{2}{7}(x^3-6y)$, which is equivalent to: $x^2+\dfrac{12y}{7} \geq \dfrac{2x}{7}(x^2+\dfrac{7y}{2})$ and since $x \leq 3$ (using Cauchy-Schwarz on the original inequality), it is sufficient to prove: $x^2+\dfrac{12y}{7} \geq \dfrac{6}{7}(x^2+\dfrac{7y}{2})$, which is equivalent to: $x^2 \geq 9y$, which follows from A.M.-G.M. ($x \geq 3y^{1/3} \geq 3y^{1/2}$) where we used $y=abc\leq 1$.

Aravind
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  • We can not assume that $a\le b\le c$ directly, right? (it is not a symmetric inequality) – River Li Jun 05 '20 at 07:04
  • Agreed. In fact, if $a \leq b \leq c$, then $a^2b+b^2c+c^2a \leq ab^2+bc^2+ca^2$, so the relevant ordering to prove the inequality is $a \leq c \leq b$. – Aravind Jun 05 '20 at 07:53
  • You have proved the case when $a \le b \le c$ (a nice solution). When $a\le c \le b$, from your result, it holds that $(a+b+c)(a+b+c - abc) \ge 2(ab^2+bc^2+ca^2)$, however, $a^2b + b^2c + c^2 a \ge ab^2+bc^2+ca^2$; Moreover, it does not hold that $(a+b+c)^3 \ge 7(a^2b+b^2c+c^2a) + 6abc$. So, you need to prove the case when $a \le c \le b$. – River Li Jun 05 '20 at 10:01
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Alternative solution:

Recall the known inequality $a^2b+b^2c+c^2a \le \frac{4}{27}(a+b+c)^3 - abc$ for $a, b, c \ge 0$. See: https://artofproblemsolving.com/community/c6h478168p2677448, https://artofproblemsolving.com/community/c4h1705936p10979968, https://artofproblemsolving.com/community/c6h479531p2685016

It suffices to prove that $(a+b+c)(a+b+c-abc) \ge \frac{8}{27}(a+b+c)^3 - 2abc$ or $$(a+b+c)^2 + 2abc \ge \frac{8}{27}(a+b+c)^3 + (a+b+c)abc.$$

By using the method of @Aravind, since $a+b+c \le 3\sqrt{\frac{a^2+b^2+c^2}{3}} \le 3$, it suffices to prove that $$(a+b+c)^2 + 2abc \ge \frac{8}{27}(a+b+c)^2 \cdot 3 + 3abc$$ or $$(a+b+c)^2 \ge 9abc.$$ Since $\frac{a+b+c}{3} \ge \sqrt[3]{abc}$ and $a+b+c\le 3$, we have $abc\le 1$ and $(a+b+c)^2 \ge 9(abc)^{2/3} \ge 9abc$. We are done.

River Li
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Another way.

We need to prove that $$(a+b+c)^2\geq2\sum_{cyc}a^2b+abc(a+b+c)$$ and since $$\frac{3}{a^2+b^2+c^2}\geq1$$ and $$1\geq\frac{a^2+b^2+c^2}{3},$$ it's enough to prove that: $$(a+b+c)^2\sqrt{\frac{a^2+b^2+c^2}{3}}\geq2\sum_{cyc}a^2b+\sqrt{\frac{3}{a^2+b^2+c^2}}abc(a+b+c)$$ or $$(a+b+c)^2(a^2+b^2+c^2)\geq2\sum_{cyc}a^2b\sqrt{3(a^2+b^2+c^2)}+3abc(a+b+c).$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$, $abc=w^3$ and $u=tv$.

Thus, $t\geq1$ and since by Rearrangement $\sum_{cyc}a^2b\leq4u^3-w^3$, it's enough to prove that: $$9u^2(9u^2-6v^2)\geq2(4u^3-w^3)\sqrt{3(9u^2-6v^2)}+9uw^3$$ or $$9u^2(3u^2-2v^2)-8u^3\sqrt{3u^2-2v^2}\geq(3u-2\sqrt{3u^2-2v^2})w^3.$$ Now, if $$(3u-2\sqrt{2u^2-2v^2}<0$$ so our inequality is obviously true because $$9u^2(3u^2-2v^2)-8u^3\sqrt{3u^2-2v^2}=$$ $$=u^2\sqrt{3u^2-2v^2}(9\sqrt{3u^2-2v^2}-8u)\geq u^2\sqrt{3u^2-2v^2}(9\sqrt{u^2}-8u)>0.$$ But for $3u-2\sqrt{2u^2-2v^2}\geq0$ since $w^3\leq v^3,$ it's enough to prove that $$9t^2(3t^2-2)-8t^3\sqrt{3t^2-2}\geq3t-2\sqrt{3t^2-2}$$ or $$27t^4-18t^2-3t\geq(8t^3-2)\sqrt{3t^2-2}$$ or $$(t-1)(537t^7+537t^6-307t^5-373t^4-49t^3-5t^2-8t-8)\geq0,$$ which is obvious for $t\geq1$.

Michael Rozenberg
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