The idea is very simple: clear denominators for the equivalent inequality \[ b^2c^2 + c^2 a^2 + a^2 b^2 \geq abc \sqrt 3 .\]Thus, it is sufficient that \[ (b^2c^2 + c^2 a^2 + a^2 b^2)^2 \geq 3 a^2b^2c^2(ab + bc + ca). \]Expanding and applying Muirhead/AM-GM finishes the problem.

HKIS200543 wrote: Give that $a,b$, and $c$ are positive real numbers such that $ab + bc + ca \geq 1$, prove that \[ \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \geq \frac{\sqrt{3}}{abc} .\] CHKMO 2019 P1? Thanks. Hong Kong (China) Mathematical Olympiad 2018

sqing wrote: CHKMO 2019 P1? Thanks. Hong Kong (China) Mathematical Olympiad 2018 It is held in December 2018 as a TST for IMO 2019.

HKIS200543 wrote: Give that $a,b$, and $c$ are positive real numbers such that $ab + bc + ca \geq 1$, prove that \[ \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \geq \frac{\sqrt{3}}{abc} .\] I will use alternative solution for this (since you already solved it): Using known inequality $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \geq \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca},$ we only need to show that: $a+b+c \geq \sqrt{3},$ which is true: $$a+b+c \geq \sqrt{3(ab+bc+ca)} \geq \sqrt{3}.$$

HKIS200543 wrote: sqing wrote: CHKMO 2019 P1? Thanks. Hong Kong (China) Mathematical Olympiad 2018 It is held in December 2018 as a TST for IMO 2019. Thanks.

HKIS200543 wrote: Give that $a,b$, and $c$ are positive real numbers such that $ab + bc + ca \geq 1$, prove that \[ \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \geq \frac{\sqrt{3}}{abc} .\] Proof of Zhangyunhua: By AM-GM, $$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} =\frac{(ab)^2+(bc)^2+(ca)^2}{(abc)^2} \geq \frac{a+b+c}{abc} \geq \frac{\sqrt{3(ab+bc+ca)}}{abc} \geq \frac{\sqrt{3}}{abc}.$$

Given that $a,b,c >1$ and $a+b+c-\sqrt{a^2+b^2+c^2}=2$, prove that$$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}< \sqrt{3}$$(BestChoice123,zither)

Allow me to use a more complicated method: \bigg(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\bigg)\big(1+1+1\big)\ge\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)^{\frac{3}{2}}\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)^{\frac{1}{2}}\ge\bigg(\frac{3}{\sqrt[3]{abc}}\bigg)^{\frac{3}{2}}\cdot\sqrt{\frac{1}{abc}}=\frac{3\sqrt3}{abc}. The above made use of Cauchy-schwarz inequality, AM-GM inequality and the given condition. (Sorry AoPS does not allow me (new user) to submit image.)

Little. Bit easy for a exam so here is my idea a very common, which is after clearing denominators we have $(ab) ^2+(bc) ^2+(ca) ^2$>=√3 ABC now on squaring both sides and using. Amgm we get the required result

TonyAoPS wrote: Allow me to use a more complicated method: $$\bigg(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\bigg)\big(1+1+1\big)\ge\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)^{\frac{3}{2}}\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)^{\frac{1}{2}}\ge\bigg(\frac{3}{\sqrt[3]{abc}}\bigg)^{\frac{3}{2}}\cdot\sqrt{\frac{1}{abc}}=\frac{3\sqrt3}{abc}.$$The above made use of Cauchy-schwarz inequality, AM-GM inequality and the given condition. (Sorry AoPS does not allow me (new user) to submit image.)

sqing wrote: Given that $a,b,c >1$ and $a+b+c-\sqrt{a^2+b^2+c^2}=2$, prove that$$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}< \sqrt{3}$$(BestChoice123,zither) $$a+b+c - \sqrt{a^2+b^2+c^2} = 2 \Longrightarrow (a+b+c-2)^2 = a^2+b^2+c^2 \Longrightarrow 2-2a-2b-2c+ab+bc+ca = 0 \Longrightarrow \boxed{c = \frac{2a+2b-2-ab}{a+b-2}}$$Also, $$c>1 \iff \frac{2a+2b-2-ab}{a+b-2} > 1 \Longrightarrow \boxed{ab < a+b}$$And, $$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}< \sqrt{3} \iff c > \frac{a+b}{\sqrt{3}ab -1} \iff \frac{2a+2b-2-ab}{a+b-2} > \frac{a+b}{\sqrt{3}ab -1}$$$$\iff -3a^2b^2+2\sqrt{3}a^2b - a^2 + 2\sqrt{3}ab^2-b^2 - 2\sqrt{3}ab-ab + 2 > 0 \text{ }(*)$$Let $a=x+1$ and $b=y+1$, then $ab<a+b \Longrightarrow xy<1$, and $(*)$ is equivalent to: $$xy(2\sqrt{3} - 1 - \sqrt{3}xy) + x^2(\sqrt{3}-1) + y^2(\sqrt{3}-1) + (2\sqrt{3}-3)(x+y) + \sqrt{3}-1 > 0$$Which is true.

HKIS200543 wrote: Given that $a,b$, and $c$ are positive real numbers such that $ab + bc + ca \geq 1$, prove that \[ \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \geq \frac{\sqrt{3}}{abc} .\] Okay, we'll clear denominators then square to get rid of the $\sqrt3$ term: $$a^4b^4+b^4c^4+c^4a^4+2a^4b^2c^2+2a^2b^4c^2+2a^2b^2c^4\ge3a^2b^2c^2.$$Homogenizing, we need to show: $$a^4b^4+b^4c^4+c^4a^4+2a^4b^2c^2+2a^2b^4c^2+2a^2b^2c^4\ge3a^2b^2c^2(ab+bc+ca)=3a^3b^3c^2+3a^2b^3c^3+3a^3b^2c^3.$$This can be done by Muirhead, summing the inequalities $(4,4,0)\succ(3,3,2)$ and twice $(4,2,2)\succ(3,3,2)$.