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Question Number 172956 by dragan91 last updated on 03/Jul/22

Commented by Shrinava last updated on 04/Jul/22

say: m=a+b+c=2 ab+ac+bc=((4−n^2 )/3)≥0 k=abc ...⇔k≤((2(n−2)(n+2))/(25+2n^2 )) ...⇔ (((2−n))/(27(25+2n^2 )))(n^2 (n^2 +(n−1)^2 )+20n^2 +2n+4)≥0 the last inequality is true since n≤2 hence proven

$$\mathrm{say}:\:\mathrm{m}=\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{2} \\ $$$$\mathrm{ab}+\mathrm{ac}+\mathrm{bc}=\frac{\mathrm{4}−\mathrm{n}^{\mathrm{2}} }{\mathrm{3}}\geqslant\mathrm{0} \\ $$$$\mathrm{k}=\mathrm{abc} \\ $$$$...\Leftrightarrow\mathrm{k}\leqslant\frac{\mathrm{2}\left(\mathrm{n}−\mathrm{2}\right)\left(\mathrm{n}+\mathrm{2}\right)}{\mathrm{25}+\mathrm{2n}^{\mathrm{2}} } \\ $$$$...\Leftrightarrow\:\frac{\left(\mathrm{2}−\mathrm{n}\right)}{\mathrm{27}\left(\mathrm{25}+\mathrm{2n}^{\mathrm{2}} \right)}\left(\mathrm{n}^{\mathrm{2}} \left(\mathrm{n}^{\mathrm{2}} +\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} \right)+\mathrm{20n}^{\mathrm{2}} +\mathrm{2n}+\mathrm{4}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{last}\:\mathrm{inequality}\:\mathrm{is}\:\mathrm{true}\:\mathrm{since}\:\mathrm{n}\leqslant\mathrm{2}\:\mathrm{hence}\:\mathrm{proven} \\ $$

Commented by Shrinava last updated on 04/Jul/22

a = b = c = (2/3)

$$\mathrm{a}\:=\:\mathrm{b}\:=\:\mathrm{c}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\: \\ $$

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