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
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 $(x,y)=(1,\sqrt 2-1)$
22.03.2023 17:20
22.03.2023 17:27
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 $y = \sqrt{2}-1, x = 1.$
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 $k\in N^+.$ $$x^2+2y\geq 2\sqrt 2-1$$$$x^3+3y\geq 3\sqrt 2-2$$
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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