Question Number 75883 by TawaTawa last updated on 19/Dec/19
$$\mathrm{If}\:\:\:\:\mathrm{a}^{\mathrm{4}} \:+\:\mathrm{b}^{\mathrm{4}} \:+\:\mathrm{c}^{\mathrm{4}} \:+\:\mathrm{d}^{\mathrm{4}} \:\:\:=\:\:\:\mathrm{16} \\ $$$$\mathrm{Prove}\:\mathrm{that},\:\:\:\:\:\:\:\:\mathrm{a}^{\mathrm{5}} \:+\:\mathrm{b}^{\mathrm{5}} \:+\:\mathrm{c}^{\mathrm{5}} \:+\:\mathrm{d}^{\mathrm{5}} \:\:\:\leqslant\:\:\:\mathrm{32} \\ $$
Commented by prakash jain last updated on 19/Dec/19
![a^4 ≤16 ⇒−2≤a≤2 a,b,c,d all lie in range [−2,2] ⇒a^5 +b^5 +c^5 +d^5 ≤2(a^4 +b^4 +c^4 +d^4 )=32■](https://www.tinkutara.com/question/Q75884.png)
$${a}^{\mathrm{4}} \leqslant\mathrm{16} \\ $$$$\Rightarrow−\mathrm{2}\leqslant{a}\leqslant\mathrm{2} \\ $$$${a},{b},{c},{d}\:\mathrm{all}\:\mathrm{lie}\:\mathrm{in}\:\mathrm{range}\:\left[−\mathrm{2},\mathrm{2}\right] \\ $$$$\Rightarrow{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} +{d}^{\mathrm{5}} \\ $$$$\leqslant\mathrm{2}\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} \right)=\mathrm{32}\blacksquare \\ $$
Commented by TawaTawa last updated on 23/Dec/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$