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Question Number 200186 by Calculusboy last updated on 15/Nov/23
Answered by ajfour last updated on 15/Nov/23
(y/x)=p , (z/y)=q , (x/z)=r =(1/(pq)) 1+p+p^2 =(2+p)((1/r))^(2/3) 1+q+q^2 =(2+q)((1/p))^(2/3) 1+r+r^2 =(2+r)((1/q))^(2/3) ......
$$\frac{{y}}{{x}}={p}\:,\:\frac{{z}}{{y}}={q}\:,\:\frac{{x}}{{z}}={r}\:=\frac{\mathrm{1}}{{pq}} \\ $$$$\mathrm{1}+{p}+{p}^{\mathrm{2}} =\left(\mathrm{2}+{p}\right)\left(\frac{\mathrm{1}}{{r}}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\mathrm{1}+{q}+{q}^{\mathrm{2}} =\left(\mathrm{2}+{q}\right)\left(\frac{\mathrm{1}}{{p}}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\mathrm{1}+{r}+{r}^{\mathrm{2}} =\left(\mathrm{2}+{r}\right)\left(\frac{\mathrm{1}}{{q}}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$…… \\ $$
Commented by witcher3 last updated on 15/Nov/23
f(x)=(((2+x)/(x+x^2 +1)))^(3/2) ;f(p)=r f(r)=q=f^2 (p) f(q)=p⇔f^3 (q)=q x≥0 f′(x)=(2/3)(((x^2 +x+1−(1+2x)(x+2))/((1+x+x^2 )^2 )))^2 (((2+x)/(x^2 +x+1)))^(1/2) =(2/3)(((−x^2 −4x−1)/((1+x+x^2 ))))(((2+x)/(x^2 +x+1)))^(1/2) <0 if f(q)>q⇒f^2 (q)<f(q) ⇒f^((3)) (q)<f^2 (q)⇒q<f^2 (q)⇒f(q)<q absurd ⇒f(q)=q (((2+x)/(x^2 +x+1)))(((2+x)/(x^2 +x+1)))^(1/2) =x (√((2+x)/(x^2 +x+1)))=x.(((x^2 +x+1)/(2+x))) if x>1 ⇒x(((x^2 +x+1)/(2+x)))=x.(1+((x(x−1))/(2+x)))>x (√((2+x)/(x^2 +x+1)))<(√((x+x^2 +x)/(x^2 +x+1)))=1 ⇒x≤1 if 0≤x<1 ⇒x(((x^2 +x+1)/(2+x)))<x(1)<1 2>x^2 +x⇒(√((2+x)/(x^2 +x+1)))>1⇒x∉IR_+ −{1} x=1 worck⇒f(q)=q⇔q=1 q=p=r=1⇔(x=y=z)
$$\mathrm{f}\left(\mathrm{x}\right)=\left(\frac{\mathrm{2}+\mathrm{x}}{\mathrm{x}+\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} ;\mathrm{f}\left(\mathrm{p}\right)=\mathrm{r} \\ $$$$\mathrm{f}\left(\mathrm{r}\right)=\mathrm{q}=\mathrm{f}^{\mathrm{2}} \left(\mathrm{p}\right) \\ $$$$\mathrm{f}\left(\mathrm{q}\right)=\mathrm{p}\Leftrightarrow\mathrm{f}^{\mathrm{3}} \left(\mathrm{q}\right)=\mathrm{q} \\ $$$$\mathrm{x}\geqslant\mathrm{0} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}−\left(\mathrm{1}+\mathrm{2x}\right)\left(\mathrm{x}+\mathrm{2}\right)}{\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\right)^{\mathrm{2}} \left(\frac{\mathrm{2}+\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{−\mathrm{x}^{\mathrm{2}} −\mathrm{4x}−\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)}\right)\left(\frac{\mathrm{2}+\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} <\mathrm{0} \\ $$$$\mathrm{if}\:\mathrm{f}\left(\mathrm{q}\right)>\mathrm{q}\Rightarrow\mathrm{f}^{\mathrm{2}} \left(\mathrm{q}\right)<\mathrm{f}\left(\mathrm{q}\right) \\ $$$$\Rightarrow\mathrm{f}^{\left(\mathrm{3}\right)} \left(\mathrm{q}\right)<\mathrm{f}^{\mathrm{2}} \left(\mathrm{q}\right)\Rightarrow\mathrm{q}<\mathrm{f}^{\mathrm{2}} \left(\mathrm{q}\right)\Rightarrow\mathrm{f}\left(\mathrm{q}\right)<\mathrm{q}\:\mathrm{absurd} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{q}\right)=\mathrm{q} \\ $$$$\left(\frac{\mathrm{2}+\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\right)\left(\frac{\mathrm{2}+\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{x} \\ $$$$\sqrt{\frac{\mathrm{2}+\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}=\mathrm{x}.\left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}{\mathrm{2}+\mathrm{x}}\right) \\ $$$$\mathrm{if}\:\mathrm{x}>\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}\left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}{\mathrm{2}+\mathrm{x}}\right)=\mathrm{x}.\left(\mathrm{1}+\frac{\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)}{\mathrm{2}+\mathrm{x}}\right)>\mathrm{x} \\ $$$$\sqrt{\frac{\mathrm{2}+\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}<\sqrt{\frac{\mathrm{x}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}\leqslant\mathrm{1}\:\:\mathrm{if}\:\mathrm{0}\leqslant\mathrm{x}<\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}\left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}{\mathrm{2}+\mathrm{x}}\right)<\mathrm{x}\left(\mathrm{1}\right)<\mathrm{1} \\ $$$$\mathrm{2}>\mathrm{x}^{\mathrm{2}} +\mathrm{x}\Rightarrow\sqrt{\frac{\mathrm{2}+\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}>\mathrm{1}\Rightarrow\mathrm{x}\notin\mathrm{IR}_{+} −\left\{\mathrm{1}\right\} \\ $$$$\mathrm{x}=\mathrm{1}\:\mathrm{worck}\Rightarrow\mathrm{f}\left(\mathrm{q}\right)=\mathrm{q}\Leftrightarrow\mathrm{q}=\mathrm{1} \\ $$$$\mathrm{q}=\mathrm{p}=\mathrm{r}=\mathrm{1}\Leftrightarrow\left(\mathrm{x}=\mathrm{y}=\mathrm{z}\right) \\ $$$$ \\ $$$$ \\ $$
Answered by AST last updated on 15/Nov/23
(2x+y)(((xzz))^(1/3) )≤(2x+y)(((x+z+z)/3))=(((x+2z)(2x+y))/3) ⇒Σ(2x+y)((xz^2 ))^(1/3) ≤((2x^2 +2y^2 +2z^2 +7xy+7zx+7zy)/3) ≤^? 2x^2 +2y^2 +2z^2 +xy+yz+zx⇔x^2 +y^2 +z^2 ≥Σxy (i),(ii)&(iii) ⇒2(x^2 +y^2 +z^2 )+xy+yz+zx=Σ(2x+y)((xz^2 ))^(1/3) But we deduced that Σ(2x+y)((xz^2 ))^(1/3) ≤2Σx^2 +Σxy Equality holds when x=y=z ⇒solution to the system={x,y,z∣x=y=z∈R^+ }
$$\left(\mathrm{2}{x}+{y}\right)\left(\sqrt[{\mathrm{3}}]{{xzz}}\right)\leqslant\left(\mathrm{2}{x}+{y}\right)\left(\frac{{x}+{z}+{z}}{\mathrm{3}}\right)=\frac{\left({x}+\mathrm{2}{z}\right)\left(\mathrm{2}{x}+{y}\right)}{\mathrm{3}} \\ $$$$\Rightarrow\Sigma\left(\mathrm{2}{x}+{y}\right)\sqrt[{\mathrm{3}}]{{xz}^{\mathrm{2}} }\leqslant\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{z}^{\mathrm{2}} +\mathrm{7}{xy}+\mathrm{7}{zx}+\mathrm{7}{zy}}{\mathrm{3}} \\ $$$$\overset{?} {\leqslant}\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{z}^{\mathrm{2}} +{xy}+{yz}+{zx}\Leftrightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \geqslant\Sigma{xy} \\ $$$$\left({i}\right),\left({ii}\right)\&\left({iii}\right) \\ $$$$\Rightarrow\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+{xy}+{yz}+{zx}=\Sigma\left(\mathrm{2}{x}+{y}\right)\sqrt[{\mathrm{3}}]{{xz}^{\mathrm{2}} } \\ $$$${But}\:{we}\:{deduced}\:{that}\:\Sigma\left(\mathrm{2}{x}+{y}\right)\sqrt[{\mathrm{3}}]{{xz}^{\mathrm{2}} }\leqslant\mathrm{2}\Sigma{x}^{\mathrm{2}} +\Sigma{xy} \\ $$$${Equality}\:{holds}\:{when}\:{x}={y}={z} \\ $$$$\Rightarrow{solution}\:{to}\:{the}\:{system}=\left\{{x},{y},{z}\mid{x}={y}={z}\in\mathbb{R}^{+} \right\} \\ $$

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