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Question Number 25025 by tawa tawa last updated on 01/Dec/17

$$\mathrm{If}{\mathrm{a}}^4+{\mathrm{b}}^4+{\mathrm{c}}^4+{\mathrm{d}}^4=16,\mathrm{prove}\mathrm{that}:{\mathrm{a}}^5+{\mathrm{b}}^5+{\mathrm{c}}^5+{\mathrm{d}}^5⩽32$$$$\mathrm{for}\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\in \mathbb{R}$$

Answered by nnnavendu last updated on 02/Dec/17

$$$$$$\mathrm{ans}$$$$$$$${\mathrm{a}}^4+{\mathrm{b}}^4+{\mathrm{c}}^4+{\mathrm{d}}^4=16$$$$\mathrm{puting}\mathrm{b}=\mathrm{c}=\mathrm{d}=0$$$${\mathrm{a}}^4+0^2+0^2+0^2=16$$$${\mathrm{a}}^4=16$$$${\mathrm{a}}^4=2^4$$$$\mathrm{a}=\mp 2$$$$\mathrm{then}$$$$$$$$\mathrm{LHS}-$$$${\mathrm{a}}^5+{\mathrm{b}}^5+{\mathrm{c}}^5+{\mathrm{d}}^4$$$${(\mp 2)}^5+0^2+0^2+0^2$$$$\mp 32\mathrm{RHS}$$$$$$

Commented by prakash jain last updated on 02/Dec/17

$$\mathrm{How}\mathrm{do}\mathrm{you}\mathrm{prove}\mathrm{that}\mathrm{for}\mathrm{any}4$$$$a,b,c\mathrm{and}d\mathrm{sum}\mathrm{will}\mathrm{remain}\mathrm{less}$$$$\mathrm{that}32.\mathrm{I}\mathrm{mean}\mathrm{the}\mathrm{general}\mathrm{case}.$$

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

$$\mathrm{\Sigma }{a}^4=16\Rightarrow a^4⩽16$$$$\Rightarrow a-1⩽1$$$$\mathrm{\Sigma }{a}^5=\mathrm{\Sigma }(a-1)a^4+\mathrm{\Sigma }{a}^4$$$$=(⩽16)+16⩽32.$$

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