Question Number 145314 by loveineq last updated on 04/Jul/21
![Let a,b ≥ 0 and (a+1)(b+1) = (a+b)^2 . Prove that (a+b)(√((a+1)^3 +(b+1)^3 )) ≤ (a+1)^2 +(b+1)^2 ≤ (1/2)[(a+1)^3 +(b+1)^3 ]](https://www.tinkutara.com/question/Q145314.png)
$$\mathrm{Let}\:{a},{b}\:\geqslant\:\mathrm{0}\:\mathrm{and}\:\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)\:=\:\left({a}+{b}\right)^{\mathrm{2}} \:.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\left({a}+{b}\right)\sqrt{\left({a}+\mathrm{1}\right)^{\mathrm{3}} +\left({b}+\mathrm{1}\right)^{\mathrm{3}} }\:\leqslant\:\left({a}+\mathrm{1}\right)^{\mathrm{2}} +\left({b}+\mathrm{1}\right)^{\mathrm{2}} \:\leqslant\:\frac{\mathrm{1}}{\mathrm{2}}\left[\left({a}+\mathrm{1}\right)^{\mathrm{3}} +\left({b}+\mathrm{1}\right)^{\mathrm{3}} \right] \\ $$
Commented by justtry last updated on 05/Jul/21
![(a+b)(b+1)=(a+b)^2 ⇔ab+a+b+1=a^2 +2b+b^2 ⇔1+(a+b)=a^2 +2b+b^2 b=0⇒(1+a)=a^2 a=0⇒(1+b)=b^2 & (1+a)^3 ≥(1+a)^2 =(a^2 )^2 (1+b)^3 ≥(1+b)^2 =(b^2 )^2 (1/2)[(1+a)^3 +(1+b)^3 ]≥(1/2)[(a^2 )^2 +(b^2 )^2 ]≥(((a^2 +b^2 )^2 )/4) ⇔(1/2)[(1+a)^3 +(1+b)^3 ]≥((a^2 )^2 +(b^2 )^2 )≥(((a^2 +b^2 )^2 )/2) ⇔ (1/2)[(1+a)^3 +(1+b)^3 ]≥(1/2)(a^4 +b^4 +2a^2 b^2 )=(1/2)[(1+a)^2 +(1+b)^2 +2(a+b)^2 ] note (1/2)[(1+a)^2 +(1+b)^2 +2(a+b)^2 ]≥(1+a)^2 +(1+b)^2 →a=b=1 or a≠b≠1, a,b≥0 so (1/2)[(1+a)^3 +(1+b)^3 ]≥(1+a)^2 +(1+b))^2 ⇔ (1+a)^2 +(1+b)^2 ≤(1/2)[(1+a)^3 +(1+b)^3 ] .........](https://www.tinkutara.com/question/Q145511.png)
$$\left({a}+{b}\right)\left({b}+\mathrm{1}\right)=\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow{ab}+{a}+{b}+\mathrm{1}={a}^{\mathrm{2}} +\mathrm{2}{b}+{b}^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{1}+\left({a}+{b}\right)={a}^{\mathrm{2}} +\mathrm{2}{b}+{b}^{\mathrm{2}} \\ $$$${b}=\mathrm{0}\Rightarrow\left(\mathrm{1}+{a}\right)={a}^{\mathrm{2}} \\ $$$${a}=\mathrm{0}\Rightarrow\left(\mathrm{1}+{b}\right)={b}^{\mathrm{2}} \\ $$$$\&\:\left(\mathrm{1}+{a}\right)^{\mathrm{3}} \geqslant\left(\mathrm{1}+{a}\right)^{\mathrm{2}} =\left({a}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\left(\mathrm{1}+{b}\right)^{\mathrm{3}} \geqslant\left(\mathrm{1}+{b}\right)^{\mathrm{2}} =\left({b}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{1}+{a}\right)^{\mathrm{3}} +\left(\mathrm{1}+{b}\right)^{\mathrm{3}} \right]\geqslant\frac{\mathrm{1}}{\mathrm{2}}\left[\left({a}^{\mathrm{2}} \right)^{\mathrm{2}} +\left({b}^{\mathrm{2}} \right)^{\mathrm{2}} \right]\geqslant\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{1}+{a}\right)^{\mathrm{3}} +\left(\mathrm{1}+{b}\right)^{\mathrm{3}} \right]\geqslant\left(\left({a}^{\mathrm{2}} \right)^{\mathrm{2}} +\left({b}^{\mathrm{2}} \right)^{\mathrm{2}} \right)\geqslant\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Leftrightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{1}+{a}\right)^{\mathrm{3}} +\left(\mathrm{1}+{b}\right)^{\mathrm{3}} \right]\geqslant\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{1}+{a}\right)^{\mathrm{2}} +\left(\mathrm{1}+{b}\right)^{\mathrm{2}} +\mathrm{2}\left({a}+{b}\right)^{\mathrm{2}} \right] \\ $$$${note} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{1}+{a}\right)^{\mathrm{2}} +\left(\mathrm{1}+{b}\right)^{\mathrm{2}} +\mathrm{2}\left({a}+{b}\right)^{\mathrm{2}} \right]\geqslant\left(\mathrm{1}+{a}\right)^{\mathrm{2}} +\left(\mathrm{1}+{b}\right)^{\mathrm{2}} \:\rightarrow{a}={b}=\mathrm{1}\:{or}\:{a}\neq{b}\neq\mathrm{1},\:{a},{b}\geqslant\mathrm{0} \\ $$$${so} \\ $$$$\left.\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{1}+{a}\right)^{\mathrm{3}} +\left(\mathrm{1}+{b}\right)^{\mathrm{3}} \right]\geqslant\left(\mathrm{1}+{a}\right)^{\mathrm{2}} +\left(\mathrm{1}+{b}\right)\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\:\left(\mathrm{1}+{a}\right)^{\mathrm{2}} +\left(\mathrm{1}+{b}\right)^{\mathrm{2}} \leqslant\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{1}+{a}\right)^{\mathrm{3}} +\left(\mathrm{1}+{b}\right)^{\mathrm{3}} \right] \\ $$$$……… \\ $$
Commented by loveineq last updated on 05/Jul/21
$$ \\ $$$$\mathrm{why}\:\mathrm{1}+{a}+{b}={a}^{\mathrm{2}} +\mathrm{2}{b}+{b}^{\mathrm{2}} ? \\ $$