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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))

1a2+1b2+1c2⩾3(abc)−23AM−GMab+bc+ac⩾⇔abc(1a+1b+1c)⩾1QM−AM⇒1a2+1b2+1c2⩾13(1a+1b+1c)2⩾131(abc)2⇒(1a2+1b2+1c2)⩾Max(13(abc)2,3(abc)23)13(abc)2⩾3(abc)23;t=abc⇒9t2⩽t23..case(1)t43⩽19t⩽133⇒abc⩽133;1abc⩾33LHS⩾13(abc)2⩾333(abc)=3abcif9t2⩾t23..case(2)⇒t⩾133,abc⩾133;(abc)13⩾13LHS⩾3(abc)23=3(abc)13abc⩾3abc3=3abcAllcaseLHS⩾3abc∀(a,b,c)∈R>03ab+ac+bc⩾1⇒1a2+1b2+1c2⩾3abc

Commented by Mingma last updated on 26/Nov/23

Very elegant, sir!

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