Let-a-b-c-0-and-a-2-b-2-c-2-3-Prove-that-a-3-b-c-3-9-

Question Number 145137 by loveineq last updated on 02/Jul/21

$$\mathrm{Let}\:{a}\geqslant{b}\geqslant{c}\geqslant\mathrm{0}\:\mathrm{and}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{3}} +\left({b}+{c}\right)^{\mathrm{3}} \:\leqslant\:\mathrm{9} \\ $$

Commented by justtry last updated on 02/Jul/21

Commented by puissant last updated on 02/Jul/21

$$\mathrm{merci} \\ $$

Commented by loveineq last updated on 02/Jul/21

$$\mathrm{is}\:\mathrm{that}\:\mathrm{proving}\:\mathrm{inequality}? \\ $$

Answered by mitica last updated on 02/Jul/21

Answered by mitica last updated on 02/Jul/21

$${f}\left({a}\right)={a}^{\mathrm{3}} +\sqrt{\left(\mathrm{6}−\mathrm{2}{a}^{\mathrm{2}} \right)^{\mathrm{3}} \:},{f}:\left[\mathrm{1},\sqrt{\mathrm{3}}\right]\rightarrow{R} \\ $$$${a}=\mathrm{1}\:{maximum}\:{point} \\ $$

Commented by loveineq last updated on 02/Jul/21

$$ \\ $$$$\mathrm{Ok},\:\mathrm{i}\:\mathrm{see}.\:\mathrm{thanks}. \\ $$

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