The following problem is from CHKMO 2018 Problem 1:
If $ab+bc+ca\ge1$, prove that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$$
I tried to use Cauchy–Schwarz inequality, by try multiplying different things, such as $1^2+1^2+1^2$, $(abc)^2+(abc)^2+(abc)^2$. But I still can’t solve it. Can someone help me?
Let $x=\dfrac{1}{a}, y=\dfrac{1}{b}, z=\dfrac{1}{c}$. It is easy to get $$x^2+y^2+z^2\ge xy+yz+zx$$ which is $$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{a+b+c}{abc}$$ We try to prove $a+b+c\ge\sqrt{3}$. As $a^2+b^2+c^2\ge ab+bc+ca\ge1$, $$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\ge3\\ a+b+c\ge\sqrt{3}$$
Because for positive variables by Muirhead we obtain: $$\sum_{cyc}a^2b^2\geq\sum_{cyc}a^2bc\geq abc\sqrt{3(ab+ac+bc)}\geq\sqrt3abc$$
AM-GM leads to \begin{align*} \frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}&\geqslant 2\sqrt{\frac{a^2b^2}{c^2}\cdot \frac{b^2c^2}{a^2}}=2b^2 \end{align*}
Similarly
\begin{align*} \frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}\geqslant 2b^2\qquad \frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}&\geqslant 2c^2\qquad \frac{c^2a^2}{b^2}+\frac{a^2b^2}{c^2}\geqslant 2a^2\\ \\ \therefore \frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}&\geqslant a^2+b^2+c^2 \end{align*}
Hence \begin{align*} \frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}+2a^2+2b^2+2c^2&\geqslant 3a^2+3b^2+3c^2\\ \iff \left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)^2&\geqslant 3a^2+3b^2+3c^2\tag{1} \end{align*}
And since $a^2+b^2+c^2\geqslant ab+bc+ac$, we obtain $$\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)^2\geqslant 3\cdot(ab+bc+ac)\geqslant3$$ Can you finish now?
Just as what was done in Post 1, we can substitute; $$a=\frac{1}{x}$$ $$b=\frac{1}{y}$$ $$c=\frac{1}{z}.$$
Then, we have that; \begin{align*} & \text{ } \text{ } \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } ab+bc+ca \geq 1\\ &\implies \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\geq 1\\ &\implies \frac{xyz}{xy}+\frac{xyz}{yz}+\frac{xyz}{zx}\geq xyz\\ &\implies x+y+z\geq xyz.\hspace{7.7cm} \rightarrow 1 \end{align*}
We can now proceed as in this link Linked Question
We know by the QM-AM inequality that; \begin{align*} \sqrt{\frac{x^2+y^2+z^2}{3}} &\geq \left(\frac{x+y+z}{3}\right)^2\\ &\geq \left(\frac{xyz}{3}\right)^2.\hspace{2cm} \left(\text{Note that }x+y+z\geq xyz\right)\\ \end{align*}
Squaring both sides and multiplying by 3 gives; $$x^2+y^2+z^2\geq \frac{x^2y^2z^2}{3}$$
\begin{align*} \implies\left(x^2+y^2+z^2\right)^\frac{1}{4} &\geq \left(\frac{x^2y^2z^2}{3}\right)^\frac{1}{4}\\ &\geq \frac{\sqrt{xyz}}{3^\frac{1}{4}}.\hspace{6cm} \rightarrow 2\\ \end{align*}
By AM-GM, we know that;
$$\frac{x^2+y^2+z^2}{3}\geq \sqrt[3]{x^2y^2z^2}$$ $$\implies x^2+y^2+z^2\geq 3\sqrt[3]{x^2y^2z^2}$$ $$\implies \left(x^2+y^2+z^2\right)^3\geq 3^3\left(x^2y^2z^2\right)$$ \begin{align*} &\implies \left(x^2+y^2+z^2\right)^\frac{3}{4}\geq 3^\frac{3}{4}\sqrt[4]{x^2y^2z^2}\\ &\implies \left(x^2+y^2+z^2\right)^\frac{3}{4}\geq 3^\frac{3}{4}\sqrt{xyz}.\hspace{5.05cm} \rightarrow 3 \end{align*}
We can now multiply Inequality 2 and 3 to have;
\begin{align*} x^2+y^2+z^2 &=\left(x^2+y^2+z^2\right)^\frac{1}{4}\times \left(x^2+y^2+z^2\right)^\frac{3}{4}\\ &\geq \frac{\sqrt{xyz}}{3^\frac{1}{4}}\times 3^\frac{3}{4}\sqrt{xyz}\\ &\geq xyz\sqrt{3}. \end{align*}
Substituting back the original variables yields the desired result; $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq\sqrt{3}\frac{1}{abc}$$ $$\implies\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq\frac{\sqrt{3}}{abc}.$$
