Question Number 82667 by M±th+et£s last updated on 23/Feb/20
$${if}\:{a}>\mathrm{0}\:\:{b}>\mathrm{0}\:\:{a}\leqslant{b} \\ $$$$ \\ $$$${show}\:{that}\: \\ $$$${a}^{\mathrm{2}} \leqslant\left(\frac{\mathrm{2}{ab}}{{a}+{b}}\right)^{\mathrm{2}} \leqslant{ab}\leqslant\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}\leqslant{b}^{\mathrm{2}} \\ $$
Answered by TANMAY PANACEA last updated on 23/Feb/20
![b^2 −((a^2 +b^2 )/2) (((b+a)(b−a))/2)>0 so b^2 >((a^2 +b^2 )/2) [equality sivn if a=b] ((a^2 +b^2 )/2)−(((a+b)/2))^2 =((2a^2 +2b^2 −a^2 −b^2 −2ab)/4)→(((b−a)^2 )/2^2 )>0 so ((a^2 +b^2 )/2)>(((a+b)/2))^2 (((a+b)/2))^2 −ab =(((a−b)^2 )/2^2 ) so (((a+b)/2))^2 >ab ab−(((2ab)/(a+b)))^2 =ab×((a^2 +2ab+b^2 −4ab)/((a+b)^2 ))→ab×(((a−b)/(a+b)))^2 so ab>(((2ab)/(a+b)))^2 (((2ab)/(a+b)))^2 −a^2 =a^2 ×((4b^2 −a^2 −2ab−b^2 )/((a+b)^2 )) (a^2 /((a+b)^2 ))×(3b^2 −3ab+ab−a^2 ) (a^2 /((a+b)^2 ))×{3b(b−a)+a(b−a)} (a^2 /((a+b)^2 ))×(b−a)(3b+a)>0 since b>a](https://www.tinkutara.com/question/Q82675.png)
$${b}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{\left({b}+{a}\right)\left({b}−{a}\right)}{\mathrm{2}}>\mathrm{0}\:\:{so}\:{b}^{\mathrm{2}} >\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}\:\:\left[{equality}\:{sivn}\:{if}\:{a}={b}\right] \\ $$$$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}−\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{2}{ab}}{\mathrm{4}}\rightarrow\frac{\left({b}−{a}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }>\mathrm{0} \\ $$$${so}\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}>\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} −{ab} \\ $$$$=\frac{\left({a}−{b}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }\:\:\:{so}\:\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} >{ab} \\ $$$${ab}−\left(\frac{\mathrm{2}{ab}}{{a}+{b}}\right)^{\mathrm{2}} \\ $$$$={ab}×\frac{{a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} −\mathrm{4}{ab}}{\left({a}+{b}\right)^{\mathrm{2}} }\rightarrow{ab}×\left(\frac{{a}−{b}}{{a}+{b}}\right)^{\mathrm{2}} \\ $$$${so}\:{ab}>\left(\frac{\mathrm{2}{ab}}{{a}+{b}}\right)^{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{2}{ab}}{{a}+{b}}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$$={a}^{\mathrm{2}} ×\frac{\mathrm{4}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} −\mathrm{2}{ab}−{b}^{\mathrm{2}} }{\left({a}+{b}\right)^{\mathrm{2}} } \\ $$$$\frac{{a}^{\mathrm{2}} }{\left({a}+{b}\right)^{\mathrm{2}} }×\left(\mathrm{3}{b}^{\mathrm{2}} −\mathrm{3}{ab}+{ab}−{a}^{\mathrm{2}} \right) \\ $$$$\frac{{a}^{\mathrm{2}} }{\left({a}+{b}\right)^{\mathrm{2}} }×\left\{\mathrm{3}{b}\left({b}−{a}\right)+{a}\left({b}−{a}\right)\right\} \\ $$$$\frac{{a}^{\mathrm{2}} }{\left({a}+{b}\right)^{\mathrm{2}} }×\left({b}−{a}\right)\left(\mathrm{3}{b}+{a}\right)>\mathrm{0}\:\:{since}\:{b}>{a} \\ $$
Commented by M±th+et£s last updated on 23/Feb/20
$${god}\:{bless}\:{you}\:{sir} \\ $$
Commented by TANMAY PANACEA last updated on 23/Feb/20
$${blessing}\:{shoser}\:{to}\:{all} \\ $$