Given $x,y>0$ such that $x^2y^2+2x^3y=1$. Find the minimum value of sum $x+y$
Problem
Source: Kazakhstan Junior National Olympiad 2023/ Problem 4
Tags: inequalities, algebra, Kazakhstan
pco
22.03.2023 17:17
rightways wrote: Given $x,y>0$ such that $x^2y^2+2x^3y=1$. Find the minimum value of sum $x+y$ $(x+y)^2=x^2+\frac{2x^3y+x^2y^2}{x^2}=x^2+\frac 1{x^2}\ge 2$ And so $\boxed{x+y\ge\sqrt 2}$, reached for example when $(x,y)=(1,\sqrt 2-1)$
Tintarn
22.03.2023 17:20
For $(1,\sqrt{2}-1)$ we get sum $\sqrt{2}$. To prove that this is minimal, it suffices to show that
\[(x+y)^4 \ge 4(x^2y^2+2x^3y).\]By homogenity it suffices to prove this for $x=1$ i.e.
\[(y+1)^4 \ge 4(y^2+2y).\]Writing $T=y^2+2y$, this reduces to
\[(T+1)^2 \ge 4T\]which is obvious.
gologinuser
22.03.2023 17:27
Let $S=x+y$ hence ${{x}^{2}}{{(S-x)}^{2}}+2{{x}^{3}}(S-x)-1=0\Leftrightarrow {{x}^{4}}-{{S}^{2}}{{x}^{2}}+1=0,x\in \mathbb{R}\Rightarrow \Delta \ge 0\Leftrightarrow S\ge \sqrt{2}$
menpo
22.03.2023 18:51
Take $f(y) = y^2\cdot x^2 + y\cdot2x^3 - 1 = 0.$ Then $$y = \dfrac{-2x^3+\sqrt{4x^6+4x^2}}{2x^2} \implies x+y = \dfrac{\sqrt{x^4+1}}{x} \ge \dfrac{\sqrt{2x^2}}{x} = \sqrt{2}.$$ As an example $y = \sqrt{2}-1, x = 1.$
sqing
23.03.2023 10:46
Let $x,y>0$ and $x^2y^2+2x^3y=1$. Prove that$$x^k+ky\geq k\sqrt 2-k+1$$Where $k\in N^+.$ $$x^2+2y\geq 2\sqrt 2-1$$$$x^3+3y\geq 3\sqrt 2-2$$
sqing
25.03.2023 12:25
rightways wrote: Given $x,y>0$ such that $x^2y^2+2x^3y=1$. Find the minimum value of sum $x+y$
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