# Prove that $\sum_{cyc}\frac{a}{b^2}\ge 3\sum_{cyc}\frac{1}{a^2}$ for $a,b,c>0$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$

$$a$$, $$b$$ and $$c$$ are three positives such that $$\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 1$$. Prove that $$\dfrac{a}{b^2} + \frac{b}{c^2} + \frac{c}{a^2} \ge 3 \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$

Here's what I did.

We have that

$$\left(\frac{a}{b^2} + \frac{b}{c^2} + \frac{c}{a^2}\right)\left(\frac{1}{b} + \frac{1}{c} + \frac{1}{a}\right) \ge \left(\sqrt{\frac{a}{b^3}} + \sqrt{\frac{b}{c^3}} + \sqrt{\frac{c}{a^3}}\right)^2$$

But because of $$\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 1$$.

$$\implies \frac{a}{b^2} + \frac{b}{c^2} + \frac{c}{a^2} \ge 3 \cdot \left(\frac{1}{b}\sqrt{\frac{a}{c^3}} + \frac{1}{c}\sqrt{\frac{b}{a^2}} + \frac{1}{a}\sqrt{\frac{c}{b^2}}\right)$$

And I am stuck, I can't think anymore.

Let $$x=\frac{1}{a}$$ , $$y=\frac{1}{b}$$ and $$z=\frac{1}{c}$$, so $$\sum_{cyc} x=1$$. Then we have

$$(LHS-RHS)\cdot xyz=(\sum_{cyc}x)\cdot(\sum_{cyc}y^3z)-3\sum_{cyc}x^3yz=\sum_{cyc}x^4y+\sum_{cyc}x^3y^2-2\sum_{cyc}x^3yz$$ $$=\frac{1}{7}\sum_{cyc}((7x^4y+4z^3x^2+2x^3y^2+y^3z^2)-14x^3yz)\geq0.$$

The last inequality come from the GM-AM by viewing $$7x^4y+4z^3x^2+2x^3y^2+y^3z^2$$ as 14 terms.

• It should be $4z^3x^2$. Nice! +1 – Michael Rozenberg Apr 30 at 7:57
• @MichaelRozenberg Yes, I have re-edited,thanks! – Bonbon Apr 30 at 8:16
• would not be easier to write $(\sum_{cyc}x)\cdot(\sum_{cyc}y^3z)$ as $(\sum_{cyc}x)\cdot(\sum_{cyc}x^3y)$, so that $x\cdot x^3y=x^4y,y\cdot x^3y=x^3y^2,z\cdot x^3y=x^3yz$? – farruhota Apr 30 at 13:25

Let $$a=\min\{a,b,c\}$$, $$b=a+u$$ and $$c=a+v$$.

Thus, $$u\geq0$$, $$v\geq0$$ and we need to prove that $$\sum_{cyc}\frac{1}{a}\sum_{cyc}\frac{a}{b^2}\geq3\sum_{cyc}\frac{1}{a^2}$$ or $$\sum_{cyc}(a^4c^3+a^4c^2b-2a^2b^3c)\geq0,$$ which is true because $$\sum_{cyc}(a^4c^3+a^4c^2b-2a^2b^3c)=4(u^2-uv+v^2)a^5+(5u^3-2u^2v+7uv^2+5v^3)a^4+$$ $$+2(u^4+3u^2v^2+6uv^3+v^4)a^3+uv(u^3+10uv^2+5v^3)a^2+2u^2v^3(u+2v)a+u^3v^3\geq0.$$

• If we are allowed to assume $abc=1$, are we also allowed to instead assume $abc = 3$? Also sorry, I deleted my previous comment to clarify some stuff because I couldn't edit it. – Darius Apr 30 at 8:46
• Yes, of course, but it does not help. – Michael Rozenberg Apr 30 at 8:47

Since the equation is homogenous, we can assume $$abc=\sqrt{3}$$ (@Michael Rozenberg).

Then,

\require{enclose} \enclose{horizontalstrike}{\begin{align*}\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2} &= \bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)\cdot\bigg(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\bigg) \\ &\geq\bigg(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\bigg)^2 \text{ by Cauchy-Schwarz}\\ &\geq\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)\cdot\bigg(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\bigg)\cdot\bigg(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\bigg)\\ &\geq\bigg(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\bigg)^2\cdot\bigg(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\bigg)\text{ by Cauchy-Schwarz}\\ &\geq\bigg(\frac{3}{abc}\bigg)^2\cdot\bigg(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\bigg)\text{ by AM-GM}\\ &\geq\bigg(\frac{3}{\sqrt{3}}\bigg)^2\cdot\bigg(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\bigg)\text{ since } abc=\sqrt{3}\\ &=3\cdot\bigg(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\bigg)\\ \end{align*}}

Edit: It's wrong.

• Sorry, but line 2 should be $(\sum_{cyc}\frac{1}{a})^2$, since $\sqrt{\frac{1}{a}\cdot\frac{a}{b^2}}=\frac{1}{b}$. – Bonbon Apr 30 at 9:09
• @Bonbon, sorry I don't understand? – Darius Apr 30 at 9:12
• I mean Cauchy-Schwarz is things like $(x^2+y^2)(z^2+w^2)\geq (xz+yw)^2$, you forget to take the square root. – Bonbon Apr 30 at 9:16
• @Bonbon oh welp :( thanks for the correction! – Darius Apr 30 at 9:18