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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))
$$\:{a},{b},{c}\:\in\mathbb{R}_{+} ^{\ast} \:\:{prove}\:{that}\:{a}+{b}+{c}\geqslant\mathrm{3}\:^{\mathrm{3}} \sqrt{{abc}} \\ $$$$ \\ $$
Commented by Matica last updated on 18/Sep/22
Please prove that AM≥GM
$${Please}\:{prove}\:{that}\:{AM}\geqslant{GM} \\ $$
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
$$\mathrm{for}\:\mathrm{2}\:\mathrm{numbers}\:\geqslant\mathrm{0} \\ $$$$\frac{{a}+{b}}{\mathrm{2}}\geqslant\sqrt{{ab}}\:\:\:\:\:\mathrm{both}\:\mathrm{sides}\:\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{we}\:\mathrm{are}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{square} \\ $$$$\frac{\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{4}}\geqslant{ab} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} \geqslant\mathrm{4}{ab} \\ $$$${a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} \geqslant\mathrm{4}{ab} \\ $$$${a}^{\mathrm{2}} −\mathrm{2}{ab}+{b}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\left({a}−{b}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\mathrm{true} \\ $$
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
$$\mathrm{for}\:\mathrm{3}\:\mathrm{numbers}\:\geqslant\mathrm{0} \\ $$$$\frac{{a}+{b}+{c}}{\mathrm{3}}\geqslant\sqrt[{\mathrm{3}}]{{abc}} \\ $$$$\frac{\left({a}+{b}+{c}\right)^{\mathrm{3}} }{\mathrm{27}}\geqslant{abc} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{3}} \geqslant\mathrm{27}{abc} \\ $$$$\mathrm{let}\:{a}\leqslant{b}\wedge{a}\leqslant{c}\wedge{p},\:{q}\geqslant\mathrm{0} \\ $$$${b}={a}+{p}\wedge{c}={a}+{q} \\ $$$$\left(\mathrm{3}{a}+{p}+{q}\right)^{\mathrm{3}} \geqslant\mathrm{27}{a}\left({a}+{p}\right)\left({a}+{q}\right) \\ $$$$\left(\mathrm{3}{a}+{p}+{q}\right)^{\mathrm{3}} −\mathrm{27}{a}\left({a}+{p}\right)\left({a}+{q}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{9}{ap}^{\mathrm{2}} −\mathrm{9}{apq}+\mathrm{9}{aq}^{\mathrm{2}} +{p}^{\mathrm{3}} +\mathrm{3}{p}^{\mathrm{2}} {q}+\mathrm{3}{pq}^{\mathrm{2}} +{q}^{\mathrm{3}} \geqslant\mathrm{0} \\ $$$$\mathrm{9}{a}\left({p}^{\mathrm{2}} −{pq}+{q}^{\mathrm{2}} \right)+\left({p}+{q}\right)^{\mathrm{3}} \geqslant\mathrm{0}\:\mathrm{true} \\ $$$${a}\geqslant\mathrm{0}\wedge\left({p}+{q}\right)^{\mathrm{3}} \geqslant\mathrm{0} \\ $$$${p}^{\mathrm{2}} −{pq}+{q}^{\mathrm{2}} =\left({p}−{q}\right)^{\mathrm{2}} +{pq}\geqslant\mathrm{0} \\ $$

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