Question-76711

Question Number 76711 by aliesam last updated on 29/Dec/19

Answered by MJS last updated on 30/Dec/19

x>0∧y>0 ⇒ x^3 +y^3 >0∧x^2 +y^2 >0 0<x^3 +y^3 ≤x−y ⇒ 0<x−y ⇔ y<x x^3 +y^3 ≤x−y let x=py∧p>1 p^3 y^3 +y^3 ≤py−y (p^3 +1)y^3 ≤(p−1)y (p^3 +1)y^2 ≤p−1 y^2 ≤((p−1)/(p^3 +1)) (p^2 +1)y^2 ≤(((p−1)(p^2 +1))/(p^3 +1)) p^2 y^2 +y^2 ≤(((p−1)(p^2 +1))/(p^3 +1)) x^2 +y^2 ≤(((p−1)(p^2 +1))/(p^3 +1)) f(p)=(((p−1)(p^2 +1))/(p^3 +1))=((p^3 −p^2 +p−1)/(p^3 +1)) f(1)=0 f′(p)=((p^4 −2p^3 +6p^2 −2p+1)/((p^3 +1)^2 ))>0∀p∈R ⇒ ⇒ p_2 >p_1 ⇔f(p_2 )>f(p_1 ) lim_(p→∞) f(p)=1 ⇒ 0<f(p)<1∀p>1 ⇒ 0<x^2 +y^2 ≤f(p)<1

$${x}>\mathrm{0}\wedge{y}>\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} >\mathrm{0}\wedge{x}^{\mathrm{2}} +{y}^{\mathrm{2}} >\mathrm{0} \\ $$$$\mathrm{0}<{x}^{\mathrm{3}} +{y}^{\mathrm{3}} \leqslant{x}−{y}\:\Rightarrow\:\mathrm{0}<{x}−{y}\:\Leftrightarrow\:{y}<{x} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} \leqslant{x}−{y} \\ $$$$\mathrm{let}\:{x}={py}\wedge{p}>\mathrm{1} \\ $$$${p}^{\mathrm{3}} {y}^{\mathrm{3}} +{y}^{\mathrm{3}} \leqslant{py}−{y} \\ $$$$\left({p}^{\mathrm{3}} +\mathrm{1}\right){y}^{\mathrm{3}} \leqslant\left({p}−\mathrm{1}\right){y} \\ $$$$\left({p}^{\mathrm{3}} +\mathrm{1}\right){y}^{\mathrm{2}} \leqslant{p}−\mathrm{1} \\ $$$${y}^{\mathrm{2}} \leqslant\frac{{p}−\mathrm{1}}{{p}^{\mathrm{3}} +\mathrm{1}} \\ $$$$\left({p}^{\mathrm{2}} +\mathrm{1}\right){y}^{\mathrm{2}} \leqslant\frac{\left({p}−\mathrm{1}\right)\left({p}^{\mathrm{2}} +\mathrm{1}\right)}{{p}^{\mathrm{3}} +\mathrm{1}} \\ $$$${p}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant\frac{\left({p}−\mathrm{1}\right)\left({p}^{\mathrm{2}} +\mathrm{1}\right)}{{p}^{\mathrm{3}} +\mathrm{1}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant\frac{\left({p}−\mathrm{1}\right)\left({p}^{\mathrm{2}} +\mathrm{1}\right)}{{p}^{\mathrm{3}} +\mathrm{1}} \\ $$$${f}\left({p}\right)=\frac{\left({p}−\mathrm{1}\right)\left({p}^{\mathrm{2}} +\mathrm{1}\right)}{{p}^{\mathrm{3}} +\mathrm{1}}=\frac{{p}^{\mathrm{3}} −{p}^{\mathrm{2}} +{p}−\mathrm{1}}{{p}^{\mathrm{3}} +\mathrm{1}} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${f}'\left({p}\right)=\frac{{p}^{\mathrm{4}} −\mathrm{2}{p}^{\mathrm{3}} +\mathrm{6}{p}^{\mathrm{2}} −\mathrm{2}{p}+\mathrm{1}}{\left({p}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} }>\mathrm{0}\forall{p}\in\mathbb{R}\:\Rightarrow \\ $$$$\Rightarrow\:{p}_{\mathrm{2}} >{p}_{\mathrm{1}} \Leftrightarrow{f}\left({p}_{\mathrm{2}} \right)>{f}\left({p}_{\mathrm{1}} \right) \\ $$$$\underset{{p}\rightarrow\infty} {\mathrm{lim}}{f}\left({p}\right)=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{0}<{f}\left({p}\right)<\mathrm{1}\forall{p}>\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\mathrm{0}<{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant{f}\left({p}\right)<\mathrm{1} \\ $$

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