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Question-200942

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Question Number 200942 by Mingma last updated on 26/Nov/23
Answered by witcher3 last updated on 26/Nov/23
(1/a^2 )+(1/b^2 )+(1/c^2 )≥3(abc)^(−(2/3)) AM−GM ab+bc+ac≥⇔ abc((1/a)+(1/b)+(1/c))≥1 QM−AM⇒ (1/a^2 )+(1/b^2 )+(1/c^2 )≥(1/3)((1/a)+(1/b)+(1/c))^2 ≥(1/3)(1/((abc)^2 )) ⇒((1/a^2 )+(1/b^2 )+(1/c^2 ))≥Max((1/(3(abc)^2 )),(3/((abc)^(2/3) ))) (1/(3(abc)^2 ))≥(3/((abc)^(2/3) ));t=abc ⇒9t^2 ≤t^(2/3) ..case(1) t^(4/3) ≤(1/9) t≤(1/(3(√3)))⇒abc≤(1/(3(√3)));(1/(abc))≥3(√3) LHS≥(1/(3(abc)^2 ))≥((3(√3))/(3(abc)))=((√3)/(abc)) if 9t^2 ≥t^(2/3) ..case (2)⇒t≥(1/(3(√3))),abc≥(1/(3(√3)));(abc)^(1/3) ≥(1/( (√3))) LHS≥(3/((abc)^(2/3) ))=((3(abc)^(1/3) )/(abc))≥ (3/(abc(√3)))=((√3)/(abc)) All case LHS≥((√3)/(abc)) ∀(a,b,c)∈R_(>0) ^3 ab+ac+bc≥1⇒ (1/a^2 )+(1/b^2 )+(1/c^2 )≥((√3)/(abc))
$$\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} }\geqslant\mathrm{3}\left(\mathrm{abc}\right)^{−\frac{\mathrm{2}}{\mathrm{3}}} \:\:\mathrm{AM}−\mathrm{GM} \\ $$$$\mathrm{ab}+\mathrm{bc}+\mathrm{ac}\geqslant\Leftrightarrow \\ $$$$\mathrm{abc}\left(\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}\right)\geqslant\mathrm{1} \\ $$$$\mathrm{QM}−\mathrm{AM}\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} }\geqslant\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}\right)^{\mathrm{2}} \geqslant\frac{\mathrm{1}}{\mathrm{3}}\frac{\mathrm{1}}{\left(\mathrm{abc}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} }\right)\geqslant\mathrm{Max}\left(\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{abc}\right)^{\mathrm{2}} },\frac{\mathrm{3}}{\left(\mathrm{abc}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} }\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{abc}\right)^{\mathrm{2}} }\geqslant\frac{\mathrm{3}}{\left(\mathrm{abc}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} };\mathrm{t}=\mathrm{abc} \\ $$$$\Rightarrow\mathrm{9t}^{\mathrm{2}} \leqslant\mathrm{t}^{\frac{\mathrm{2}}{\mathrm{3}}} ..\mathrm{case}\left(\mathrm{1}\right) \\ $$$$\mathrm{t}^{\frac{\mathrm{4}}{\mathrm{3}}} \leqslant\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\mathrm{t}\leqslant\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}\Rightarrow\mathrm{abc}\leqslant\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}};\frac{\mathrm{1}}{\mathrm{abc}}\geqslant\mathrm{3}\sqrt{\mathrm{3}} \\ $$$$\mathrm{LHS}\geqslant\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{abc}\right)^{\mathrm{2}} }\geqslant\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{3}\left(\mathrm{abc}\right)}=\frac{\sqrt{\mathrm{3}}}{\mathrm{abc}} \\ $$$$\mathrm{if}\:\mathrm{9t}^{\mathrm{2}} \geqslant\mathrm{t}^{\frac{\mathrm{2}}{\mathrm{3}}} ..\mathrm{case}\:\left(\mathrm{2}\right)\Rightarrow\mathrm{t}\geqslant\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}},\mathrm{abc}\geqslant\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}};\left(\mathrm{abc}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \geqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{LHS}\geqslant\frac{\mathrm{3}}{\left(\mathrm{abc}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} }=\frac{\mathrm{3}\left(\mathrm{abc}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{abc}}\geqslant\:\frac{\mathrm{3}}{\mathrm{abc}\sqrt{\mathrm{3}}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{abc}} \\ $$$$\mathrm{All}\:\mathrm{case}\:\mathrm{LHS}\geqslant\frac{\sqrt{\mathrm{3}}}{\mathrm{abc}} \\ $$$$\forall\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)\in\mathbb{R}_{>\mathrm{0}} ^{\mathrm{3}} \\ $$$$\mathrm{ab}+\mathrm{ac}+\mathrm{bc}\geqslant\mathrm{1}\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} }\geqslant\frac{\sqrt{\mathrm{3}}}{\mathrm{abc}} \\ $$
Commented by Mingma last updated on 26/Nov/23
Very elegant, sir!

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