Menu Close

if-0-x-y-z-k-and-k-gt-0-then-y-x-z-z-x-k-k-2-




Question Number 151100 by mathdanisur last updated on 18/Aug/21
if  0≤x;y;z≤k  and  k>0  then:  y(x - z) - z(x - k) ≤ k^2
$$\mathrm{if}\:\:\mathrm{0}\leqslant\mathrm{x};\mathrm{y};\mathrm{z}\leqslant\mathrm{k}\:\:\mathrm{and}\:\:\mathrm{k}>\mathrm{0}\:\:\mathrm{then}: \\ $$$$\mathrm{y}\left(\mathrm{x}\:-\:\mathrm{z}\right)\:-\:\mathrm{z}\left(\mathrm{x}\:-\:\mathrm{k}\right)\:\leqslant\:\mathrm{k}^{\mathrm{2}} \\ $$
Answered by dumitrel last updated on 18/Aug/21
⇔y(x−z)+z(k−x)≤k^2   I.  If x≤z⇒y(x−z)≤0         z(k−x)≤k∙k=k^2 ⇒        ⇒y(x−z)+z(k−x)≤k^2   II.  Ifx>z⇒         ∣y(x−z)+z(k−x)∣≤∣y(x−z)∣+∣z(k−x)∣≤         ≤k(x−z)+k(k−x)=k(k−z)≤k^2    ⇒y(x−z)+z(k−x)≤k^2
$$\Leftrightarrow{y}\left({x}−{z}\right)+{z}\left({k}−{x}\right)\leqslant{k}^{\mathrm{2}} \\ $$$${I}.\:\:{If}\:{x}\leqslant{z}\Rightarrow{y}\left({x}−{z}\right)\leqslant\mathrm{0} \\ $$$$\:\:\:\:\:\:\:{z}\left({k}−{x}\right)\leqslant{k}\centerdot{k}={k}^{\mathrm{2}} \Rightarrow \\ $$$$\:\:\:\:\:\:\Rightarrow{y}\left({x}−{z}\right)+{z}\left({k}−{x}\right)\leqslant{k}^{\mathrm{2}} \\ $$$${II}.\:\:{Ifx}>{z}\Rightarrow \\ $$$$\:\:\:\:\:\:\:\mid{y}\left({x}−{z}\right)+{z}\left({k}−{x}\right)\mid\leqslant\mid{y}\left({x}−{z}\right)\mid+\mid{z}\left({k}−{x}\right)\mid\leqslant \\ $$$$\:\:\:\:\:\:\:\leqslant{k}\left({x}−{z}\right)+{k}\left({k}−{x}\right)={k}\left({k}−{z}\right)\leqslant{k}^{\mathrm{2}} \\ $$$$\:\Rightarrow{y}\left({x}−{z}\right)+{z}\left({k}−{x}\right)\leqslant{k}^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mathdanisur last updated on 18/Aug/21
nice Ser thank you
$$\mathrm{nice}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{thank}\:\mathrm{you} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *