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Question Number 150903 by puissant last updated on 16/Aug/21

let x,y>0 , n∈N,  show that (x+y)^n ≤2^(n−1) (x^n +y^n )..

$${let}\:{x},{y}>\mathrm{0}\:,\:{n}\in\mathbb{N}, \\ $$ $${show}\:{that}\:\left({x}+{y}\right)^{{n}} \leqslant\mathrm{2}^{{n}−\mathrm{1}} \left({x}^{{n}} +{y}^{{n}} \right).. \\ $$

Answered by qaz last updated on 16/Aug/21

Consider strickly ↗ or ↘ convex fuction f(x)=x^n        ,(x>0)  have f(((x+y)/2))≤(1/2)(f(x)+f(y))  ie.(((x+y)/2))^n ≤(1/2)(x^n +y^n )  ⇒(x+y)^n ≤2^(n−1) (x^n +y^n )

$$\mathrm{Consider}\:\mathrm{strickly}\:\nearrow\:\mathrm{or}\:\searrow\:\mathrm{convex}\:\mathrm{fuction}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{n}} \:\:\:\:\:\:\:,\left(\mathrm{x}>\mathrm{0}\right) \\ $$ $$\mathrm{have}\:\mathrm{f}\left(\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\right)\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}\right)\right) \\ $$ $$\mathrm{ie}.\left(\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\right)^{\mathrm{n}} \leqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}^{\mathrm{n}} +\mathrm{y}^{\mathrm{n}} \right) \\ $$ $$\Rightarrow\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{n}} \leqslant\mathrm{2}^{\mathrm{n}−\mathrm{1}} \left(\mathrm{x}^{\mathrm{n}} +\mathrm{y}^{\mathrm{n}} \right) \\ $$

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