Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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 bymathdanisur last updated on 18/Aug/21

nice Ser thank you

$$\mathrm{nice}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{thank}\:\mathrm{you} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com