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(a)=a^3 +(√((6−2a^2 )^3 )),f:[1,(√3)]→R a=1 maximum point](Q145146.png)
$${f}\left({a}\right)={a}^{\mathrm{3}} +\sqrt{\left(\mathrm{6}−\mathrm{2}{a}^{\mathrm{2}} \right)^{\mathrm{3}} \:},{f}:\left[\mathrm{1},\sqrt{\mathrm{3}}\right]\rightarrow{R} \\ $$$${a}=\mathrm{1}\:{maximum}\:{point} \\ $$
Commented by loveineq last updated on 02/Jul/21
$$ \\ $$$$\mathrm{Ok},\:\mathrm{i}\:\mathrm{see}.\:\mathrm{thanks}. \\ $$