$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}$

Question

Prove the following inequality $$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}, \enspace \forall a,b,c \in (0,\infty)$$

I tried by multying both sides by the denominator $(a+b+c)^2$ and then applying Holder for the left side but I couldn’t work it out. I would prefer a proof without never ending computations (a.k.a. Brute Force / opening up the brackets) because I know how to do it this way already. A proof using C-B-S, Holder, Titu’s Lemma or their generalizations or other well-known inequalities would be ideal. Thank you!

Answer 1

This seems to be deceptively cumbersome, needing a tight bound on the LHS while symmetrisation. The following is from my old notes, unfortunately no way to attribute it correctly....

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First we need a well known inequality $4(x+y+z)^3 \geqslant 27(x^2y+y^2z+z^2x+xyz)$, which follows from AM-GM, as WLOG we may assume $y$ is in between $x, z$: $$\frac{\frac{x+z}2+y+\frac{x+z}2}3\geqslant \sqrt[3]{\frac{x+z}2\cdot y \cdot \frac{x+z}2}$$ $$\implies \frac4{27}(x+y+z)^3 = x^2y+y^2z+z^2x+xyz - z(y-x)(y-z) \\\geqslant (x^2y+y^2z+z^2x+xyz)$$

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Using the above with $x = \frac{a}b, y=\frac{b}c, z = \frac{c}a$, we get $$\left(\frac{a}b+\frac{b}c+\frac{c}a \right)^3 \geqslant \frac{27}4\left( \frac{a^3+b^3+c^3}{abc}+1\right)$$

Hence to prove the original inequality, it is enough to show instead the symmetric $$\frac{a^3+b^3+c^3}{abc} +1\geqslant 108 \frac{(a^2+b^2+c^2)^3}{(a+b+c)^6} \tag{1}$$

$$\iff \frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}{abc}+4\geqslant \frac{108(a^2+b^2+c^2)^3}{(a+b+c)^6} \tag{2}$$

Using the obvious $(ab+bc+ca)^2 \geqslant 3abc(a+b+c)$, $$\frac{a+b+c}{abc} \geqslant \frac{3(a+b+c)^2}{(ab+bc+ac)^2}$$ and hence, it is enough to show $$\frac{3(a+b+c)^2(a^2+b^2+c^2-ab-bc-ac)}{(ab+bc+ac)^2} + 4 \geqslant \frac{108(a^2+b^2+c^2)^3}{(a+b+c)^6} \tag{3}$$ Setting $t=\dfrac{3(a^2+b^2+c^2)}{(a+b+c)^2}\, \in [1, 3)$ we need to equivalently show $$\frac{54(t-1)}{(3-t)^2}+4\geqslant 4t^3 \tag{4}$$ which is $(t-1)(9-6t-8t^2+10t^3-2t^4)\geqslant 0$ and hence holds true as $$9-6t-8t^2+10t^3-2t^4=2(3+3t-t^2)(t-1)^2+3>0$$

Answer 2

My old proof.

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

Thus, $t\geq1$ and we need to prove that: $$u^2\sum_{cyc}a^2c\geq3(3u^2-2v^2)w^3$$ or $$u^2\sum_{cyc}2a^2c\geq6(3u^2-2v^2)w^3$$ or $$u^2\sum_{cyc}(a^2b+a^2c)-6(3u^2-2v^2)w^3\geq u^2\sum_{cyc}(a^2b-a^2c)$$ or $$u^2(9uv^2-3w^3)-6(3u^2-2v^2)w^3\geq u^2(a-b)(a-c)(b-c)$$ or $$3(3u^3v^2-7u^2w^3+4v^2w^3)\geq u^2(a-b)(a-c)(b-c).$$ We'll prove that $$3u^3v^2-7u^2w^3+4v^2w^3\geq0.$$

Indeed, $$(a-b)^2(a-c)^2(b-c)^2\geq0$$ gives $$27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)\geq0.$$ Hence, $$w^3\leq3uv^2-2u^3+2\sqrt{(u^2-v^2)^3}$$ and it's enough to prove that $$3u^3v^2\geq(7u^2-4v^2)\left(3uv^2-2u^3+2\sqrt{(u^2-v^2)^3}\right)$$ or $$7u^5-13u^3v^2+6uv^4\geq(7u^2-4v^2)\sqrt{(u^2-v^2)^3}$$ or $$7u^3-6uv^2\geq(7u^2-4v^2)\sqrt{u^2-v^2}$$ or $$(21u^4-36u^2v^2+16v^4)v^2\geq0,$$ which is obvious.

Thus, it remains to prove that: $$(3u^3v^2-7u^2w^3+4v^2w^3)^2\geq3u^4(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6),$$ which is $$(13u^4-14u^2v^2+4v^4)w^6+3(u^4-5u^2v^2+2v^4)u^3w^3+3u^4v^6\geq0.$$ Now, if $u^4-5u^2v^2+2v^4\geq0$ so the last inequality is obviously true.

Hence, it's enough to prove our inequality for $u^4-5u^2v^2+2v^4<0$ or $1\leq t<\frac{5+\sqrt{17}}{2}$.

Id est, it's enough to prove that $$3u^6(u^4-5u^2v^2+2v^4)^2-4(13u^4-14u^2v^2+4v^4)u^4v^6\leq0$$ or $$3t(t^2-5t+2)^2-4(13t^2-14t+4)\leq0$$ or $$(t-1)^2(3t^3-24t^2+36t-16)\leq0,$$ which is true for $1\leq t<\frac{5+\sqrt{17}}{2}$.

Done!

Answer 3

There is also the following proof (not mine) by SOS. $$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a}-\frac{9(a^2+b^2+c^2)}{(a+b+c)^2}=\frac{\sum\limits_{cyc}\left((a^2-ab)abc+a(a-b)^2(b-2c)^2\right)}{abc(a+b+c)^2}\geq0.$$

Answer 4

There is also another following proof (not mine) by SOS, it is nice although I can't understand how to get it.

$$\text{LHS-RHS} = \frac{1}{16(a+b+c)^2}\sum \frac{[15c^2 +(4b-7c)^2](a-b)^2}{bc}$$