Question Number 25025 by tawa tawa last updated on 01/Dec/17
$$\mathrm{If}\:\:\:\:\mathrm{a}^{\mathrm{4}} \:+\:\mathrm{b}^{\mathrm{4}} \:+\:\mathrm{c}^{\mathrm{4}} \:+\:\mathrm{d}^{\mathrm{4}} \:=\:\mathrm{16},\:\:\mathrm{prove}\:\mathrm{that}:\:\:\mathrm{a}^{\mathrm{5}} \:+\:\mathrm{b}^{\mathrm{5}} \:+\:\mathrm{c}^{\mathrm{5}} \:+\:\mathrm{d}^{\mathrm{5}} \:\leqslant\:\mathrm{32} \\ $$$$\mathrm{for}\:\:\mathrm{a},\:\mathrm{b},\:\mathrm{c},\:\mathrm{d}\:\in\:\mathbb{R} \\ $$
Answered by nnnavendu last updated on 02/Dec/17
$$ \\ $$$$\mathrm{ans} \\ $$$$ \\ $$$$\mathrm{a}^{\mathrm{4}} +\mathrm{b}^{\mathrm{4}} +\mathrm{c}^{\mathrm{4}} +\mathrm{d}^{\mathrm{4}} =\mathrm{16} \\ $$$$\mathrm{puting}\:\:\mathrm{b}=\mathrm{c}=\mathrm{d}=\mathrm{0} \\ $$$$\mathrm{a}^{\mathrm{4}} +\mathrm{0}^{\mathrm{2}} +\mathrm{0}^{\mathrm{2}} +\mathrm{0}^{\mathrm{2}} =\mathrm{16} \\ $$$$\mathrm{a}^{\mathrm{4}} =\mathrm{16} \\ $$$$\mathrm{a}^{\mathrm{4}} =\mathrm{2}^{\mathrm{4}} \\ $$$$\mathrm{a}=\mp\mathrm{2} \\ $$$$\mathrm{then} \\ $$$$ \\ $$$$\mathrm{LHS}− \\ $$$$\mathrm{a}^{\mathrm{5}} +\mathrm{b}^{\mathrm{5}} +\mathrm{c}^{\mathrm{5}} +\mathrm{d}^{\mathrm{4}} \\ $$$$\left(\mp\mathrm{2}\right)^{\mathrm{5}} +\mathrm{0}^{\mathrm{2}} +\mathrm{0}^{\mathrm{2}} +\mathrm{0}^{\mathrm{2}} \\ $$$$\mp\mathrm{32}\:\:\:\mathrm{RHS} \\ $$$$ \\ $$
Commented by prakash jain last updated on 02/Dec/17
$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{any}\:\mathrm{4} \\ $$$${a},{b},{c}\:\mathrm{and}\:{d}\:\mathrm{sum}\:\mathrm{will}\:\mathrm{remain}\:\mathrm{less} \\ $$$$\mathrm{that}\:\mathrm{32}.\:\mathrm{I}\:\mathrm{mean}\:\mathrm{the}\:\mathrm{general}\:\mathrm{case}. \\ $$
Commented by tawa tawa last updated on 10/Dec/17
Thanks for your help.
Commented by tawa tawa last updated on 10/Dec/17
God bless you sir
Answered by ajfour last updated on 02/Dec/17
$$\:\Sigma{a}^{\mathrm{4}} =\mathrm{16}\:\:\:\:\Rightarrow\:\:\:{a}^{\mathrm{4}} \leqslant\:\mathrm{16} \\ $$$$\Rightarrow\:\:{a}−\mathrm{1}\:\leqslant\:\mathrm{1} \\ $$$$\Sigma{a}^{\mathrm{5}} =\:\Sigma\left({a}−\mathrm{1}\right){a}^{\mathrm{4}} +\Sigma{a}^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\left(\leqslant\:\mathrm{16}\right)+\mathrm{16}\:\leqslant\:\mathrm{32}\:. \\ $$