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Question Number 205420 by hardmath last updated on 20/Mar/24

$$\mathrm{If}$$$${\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{a}}^2+{\mathrm{b}}^2=4$$$$\mathrm{Then}:$$$${\mathrm{a}}^4+{\mathrm{b}}^4⩾2$$

Answered by Berbere last updated on 21/Mar/24

$$2(a^4+b^4)+a^2+b^2+1+1$$$$=a^4+a^2+b^4+b^2+a^4+1+b^4+1$$$$\stackrel{AM-GM}{⩾}2\mid a^3\mid +2\mid b^3\mid +2a^2+2b^2⩾2(a^3+b^3+a^2+b^2)=8$$$$2(a^4+b^4)+a^2+b^2⩾6$$$$a^4+b^4\stackrel{AM-QM}{⩾}\frac{{(a^2+b^2)}^2}{2};X=a^2+b^2⩾0$$$$a^4+b^4⩾Max(\frac{6-X}{2},\frac{X^2}{2})=\frac{1}{2}Max(6-X,X^2)$$$$X^2+X-6=(X+3)(X-2)$$$$X\in [-3,2]max(\frac{X^2}{2},\frac{1}{2}(6-X))=\frac{1}{2}(6-X)⩾\frac{6-2}{2}=2$$$$X⩽-3;Max(\frac{x^2}{2},\frac{6-x}{2})=\frac{{(-3)}^2}{2}⩾2$$$$X⩾2Max(\frac{X^2}{2},\frac{6-X}{2})=\frac{X^2}{2}⩾2$$$$\Rightarrow \mathrm{\forall }X=a^2+b^2\in [0,\mathrm{\infty }[a^4+b^4⩾2$$$$$$$$$$$$$$

Commented by hardmath last updated on 21/Mar/24

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}\mathrm{cool}$$

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