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Question Number 176399 by Matica last updated on 18/Sep/22

a,b,c ∈R_+ ^∗ prove that a+b+c≥3^3 (√(abc))

a,b,c∈R+∗provethata+b+c⩾33abc

Commented by Matica last updated on 18/Sep/22

Please prove that AM≥GM

PleaseprovethatAM⩾GM

Answered by Frix last updated on 18/Sep/22

for 2 numbers ≥0 ((a+b)/2)≥(√(ab)) both sides ≥0 ⇒ we are allowed to square (((a+b)^2 )/4)≥ab (a+b)^2 ≥4ab a^2 +2ab+b^2 ≥4ab a^2 −2ab+b^2 ≥0 (a−b)^2 ≥0 true

for2numbers⩾0a+b2⩾abbothsides⩾0⇒weareallowedtosquare(a+b)24⩾ab(a+b)2⩾4aba2+2ab+b2⩾4aba2−2ab+b2⩾0(a−b)2⩾0true

Answered by Frix last updated on 18/Sep/22

for 3 numbers ≥0 ((a+b+c)/3)≥((abc))^(1/3) (((a+b+c)^3 )/(27))≥abc (a+b+c)^3 ≥27abc let a≤b∧a≤c∧p, q≥0 b=a+p∧c=a+q (3a+p+q)^3 ≥27a(a+p)(a+q) (3a+p+q)^3 −27a(a+p)(a+q)≥0 9ap^2 −9apq+9aq^2 +p^3 +3p^2 q+3pq^2 +q^3 ≥0 9a(p^2 −pq+q^2 )+(p+q)^3 ≥0 true a≥0∧(p+q)^3 ≥0 p^2 −pq+q^2 =(p−q)^2 +pq≥0

for3numbers⩾0a+b+c3⩾abc3(a+b+c)327⩾abc(a+b+c)3⩾27abcleta⩽b∧a⩽c∧p,q⩾0b=a+p∧c=a+q(3a+p+q)3⩾27a(a+p)(a+q)(3a+p+q)3−27a(a+p)(a+q)⩾09ap2−9apq+9aq2+p3+3p2q+3pq2+q3⩾09a(p2−pq+q2)+(p+q)3⩾0truea⩾0∧(p+q)3⩾0p2−pq+q2=(p−q)2+pq⩾0

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