a, ab + bc + ca = 3
$<=> \frac{ab + bc +ca}{abc}= \frac{3}{abc}$
$<=> \frac{1}{a}+\frac{1}{b}+\frac{1}{c}= \frac{3}{abc}$
$<=> \left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^{2}= \frac{9}{a^{2}b^{2}c^{2}}$
Ta có $\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^{2}\leq 3\left ( \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \right )$
=> $3\left ( \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \right )\geq \frac{9}{a^{2}b^{2}c^{2}}$
$<=> \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c2}\geq \frac{3}{a^{2}b^{2}c^{2}}$
Mặt khác $ab + bc + ca \geq 3\sqrt[3]{a^{2}b^{2}c^{2}}
<=> 3 \geq 3\sqrt[3]{a^{2}b^{2}c^{2}}$$<=> 1 \geq \sqrt[3]{a^{2}b^{2}c^{2}}
<=> 1 \geq a^{2}b^{2}c^{2} =>\frac{3}{a^{2}b{2}c^{2}}\geq 3=>...$