Prove $(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$

Question

For $a,b,c>0; abc=1.$ Prove$:$ $$(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$$

My proof by SOS is ugly and hard if without computer$:$

$$\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{3}-9\,abc \left( {a}^{3}+{b} ^{3}+{c}^{3} \right)$$

$$=\frac{1}{8}\, \left( b-c \right) ^{6}+{\frac {117\, \left( b+c \right) ^{4} \left( b+c-2\,a \right) ^{2}}{1024}}+{\frac {3\,{a}^{2} \left( 40\,{a }^{2}+7\,{b}^{2}+14\,bc+7\,{c}^{2} \right) \left( b-c \right) ^{2}}{ 32}}$$

$$+{\frac {3\, \left( b+c \right) ^{2} \left( 3\,a-2\,b-2\,c \right) ^{2} \left( b-c \right) ^{2}}{32}}+\frac{3}{16}\, \left( a+2\,b+2\,c \right) \left( 4\,a+b+c \right) \left( b-c \right) ^{4}$$

$$+{\frac { \left( 16\,{a}^{2}+24\,ab+24\,ac+11\,{b}^{2}+22\,bc+11\,{c}^{ 2} \right) \left( 4\,a-b-c \right) ^{2} \left( b+c-2\,a \right) ^{2} }{1024}} \geqq 0$$

I think$,$ $uvw$ is the best way here but it's not concordant for student in The Secondary School.

Also$,$ BW helps here, but not is nice, I think.

So I wanna nice solution for it! Thanks for a real lot!

Answer 1

Yes, SOS helps: $$(a^2+b^2+c^2)^3-9(a^3+b^3+c^3)=(a^2+b^2+c^2)^3-9abc(a^3+b^3+c^3)=$$ $$=\frac{1}{2}\sum_{cyc}(2a^6+6a^4b^2+6a^4c^2-18a^4bc+4a^2b^2c^2)=$$ $$=\frac{1}{2}\sum_{cyc}(2a^6-a^4b^2-a^4c^2+7a^4b^2+7a^4c^2-14c^4ab-4a^4bc+4a^2b^2c^2)=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)^2((a+b)^2(a^2+b^2)+7c^4-2abc(a+b+c))=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)^2(7c^4-2abc^2-2ab(a+b)c+(a+b)^2(a^2+b^2))\geq0,$$ where the last inequality is true by AM-GM: $$c^4+\frac{1}{8}(a^2+b^2)(a+b)^2\geq c^4+a^2b^2\geq2abc^2$$ and $$6c^4+\frac{7}{8}(a+b)^2(a^2+b^2)\geq2ab(a+b)c.$$ Can you prove the last inequality by AM-GM?

Answer 2

Because $(a+b+c)(ab+bc+ca) \geqslant 9abc,$ so we will prove stronger inequality $$(a^2+b^2+c^2)^3 \geqslant (a+b+c)(ab+bc+ca)(a^3+b^3+c^3).$$ or $$(a^2+b^2+c^2-ab-bc-ca)^2\sum (a^2+bc)+ \frac{ab+bc+ca}{2} \sum a^2(b-c)^2 \geqslant 0.$$ Done.

Answer 3

Suppose $a = \max\{a,b,c\}.$ By the AM-GM inequality we have $$9abc(a^3+b^3+c^3) \leqslant \left(ab+ca+\frac{a^3+b^3+c^3}{3a}\right)^3.$$ Therefore, we need to prove $$a^2+b^2+c^2 \geqslant ab+ca+\frac{a^3+b^3+c^3}{3a},$$ equivalent to $$\frac{(2a-b-c)(a^2+b^2+c^2-ab-bc-ca)}{3a} \geqslant 0.$$ which is true.

Answer 4

By applying Vacs's ineq we obtain $$(a^2+b^2+c^2)^3\ge 3(ab^3+bc^3+ca^3)(a^2+b^2+c^2)$$ So it's suffice to prove $$(ab^3+bc^3+ca^3)(a^2+b^2+c^2)\ge 3abc(a^3+b^3+c^3)$$ $$\Leftrightarrow \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge \frac{3(a^3+b^3+c^3)}{a^2+b^2+c^2}$$ $$\Leftrightarrow (a-b)^2\left(\frac{a^2+c^2}{b}-a\right)+(b-c)^2\left(\frac{a^2+b^2}{c}-b\right)+(c-a)^2\left(\frac{b^2+c^2}{a}-c\right)$$ $$\Leftrightarrow S=S_c(a-b)^2+S_a(b-c)^2+S_b(c-a)^2\ge 0$$ Assume $b=\text{mid} \{a,b,c\}$ we consider 2 cases

Case 1: $a\ge b\ge c\implies S_a,S_c\ge 0$ and $S_a+2S_b, \,\ S_c+2S_b\ge 0\implies S\ge 0$

Case 2: $c\ge b\ge a\implies S_b,\,\ S_c,\,\ S_a+S_b\ge 0\implies S\ge 0$ So we are done