Question Number 205420 by hardmath last updated on 20/Mar/24
$$\mathrm{If} \\ $$$$\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \:+\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:=\:\mathrm{4} \\ $$$$\mathrm{Then}: \\ $$$$\mathrm{a}^{\mathrm{4}} \:+\:\mathrm{b}^{\mathrm{4}} \:\geqslant\:\mathrm{2} \\ $$
Answered by Berbere last updated on 21/Mar/24
![2(a^4 +b^4 )+a^2 +b^2 +1+1 =a^4 +a^2 +b^4 +b^2 +a^4 +1+b^4 +1 ≥^(AM−GM) 2∣a^3 ∣+2∣b^3 ∣+2a^2 +2b^2 ≥2(a^3 +b^3 +a^2 +b^2 )=8 2(a^4 +b^4 )+a^2 +b^2 ≥6 a^4 +b^4 ≥^(AM−QM) (((a^2 +b^2 )^2 )/2);X=a^2 +b^2 ≥0 a^4 +b^4 ≥Max(((6−X)/2),(X^2 /2))=(1/2)Max(6−X,X^2 ) X^2 +X−6=(X+3)(X−2) X∈[−3,2] max((X^2 /2),(1/2)(6−X))=(1/2)(6−X)≥((6−2)/2)=2 X≤−3; Max((x^2 /2),((6−x)/2))=(((−3)^2 )/2)≥2 X≥2 Max((X^2 /2),((6−X)/2))=(X^2 /2)≥2 ⇒∀X=a^2 +b^2 ∈[0,∞[ a^4 +b^4 ≥2](https://www.tinkutara.com/question/Q205447.png)
$$\mathrm{2}\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} \right)+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{1}+\mathrm{1} \\ $$$$={a}^{\mathrm{4}} +{a}^{\mathrm{2}} +{b}^{\mathrm{4}} +{b}^{\mathrm{2}} +{a}^{\mathrm{4}} +\mathrm{1}+{b}^{\mathrm{4}} +\mathrm{1} \\ $$$$\overset{{AM}−{GM}} {\geqslant}\mathrm{2}\mid{a}^{\mathrm{3}} \mid+\mathrm{2}\mid{b}^{\mathrm{3}} \mid+\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} \geqslant\mathrm{2}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\mathrm{8} \\ $$$$\mathrm{2}\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} \right)+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \geqslant\mathrm{6} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} \overset{{AM}−{QM}} {\geqslant}\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}};{X}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} \geqslant{Max}\left(\frac{\mathrm{6}−{X}}{\mathrm{2}},\frac{{X}^{\mathrm{2}} }{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}{Max}\left(\mathrm{6}−{X},{X}^{\mathrm{2}} \right) \\ $$$${X}^{\mathrm{2}} +{X}−\mathrm{6}=\left({X}+\mathrm{3}\right)\left({X}−\mathrm{2}\right)\:\: \\ $$$${X}\in\left[−\mathrm{3},\mathrm{2}\right]\:\:{max}\left(\frac{{X}^{\mathrm{2}} }{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{6}−{X}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{6}−{X}\right)\geqslant\frac{\mathrm{6}−\mathrm{2}}{\mathrm{2}}=\mathrm{2} \\ $$$${X}\leqslant−\mathrm{3};\:{Max}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}},\frac{\mathrm{6}−{x}}{\mathrm{2}}\right)=\frac{\left(−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{2}}\geqslant\mathrm{2} \\ $$$${X}\geqslant\mathrm{2}\:\:{Max}\left(\frac{{X}^{\mathrm{2}} }{\mathrm{2}},\frac{\mathrm{6}−{X}}{\mathrm{2}}\right)=\frac{{X}^{\mathrm{2}} }{\mathrm{2}}\geqslant\mathrm{2} \\ $$$$\Rightarrow\forall{X}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \in\left[\mathrm{0},\infty\left[\:\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} \geqslant\mathrm{2}\right.\right. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by hardmath last updated on 21/Mar/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool} \\ $$