Very easy inequality

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Very easy inequality

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Source: Polish Junior MO Second Round 2019

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pggp

#1 h Oct 26, 2020, 1:52 PM

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Let $x$ , $y$ be real numbers, such that $x^2 + x \leq y$ . Prove that $y^2 + y \geq x$ .

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#2 h Oct 26, 2020, 3:07 PM • 2 Y

Y by Mango247, pavel kozlov

$y^2+y\ge (x^2+x)^2+(x^2+x)= ((x^2+x)^2+x^2)+x\ge x$ since squares of reals are greater than zero.

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#3 h Mar 19, 2025, 6:06 PM

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Easy just notice that $y^2+y \ge y \ge x^2+x \ge x$ :cool:

:cool:

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#4 h an hour ago

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pggp wrote:

Let $x$ , $y$ be real numbers, such that $x^2 + x \leq y$ . Prove that $y^2 + y \geq x$ .

We prove by contradiction method the following proposition
Asume that there is a and real value of x and y such that
${{y}^{2}}+y<x\Rightarrow {{y}^{2}}+y+{{x}^{2}}+x<x+y\Rightarrow {{x}^{2}}+{{y}^{2}}<0$ ,is false

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#5 h an hour ago

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Let $x$ , $y$ be real numbers and $a,b>0$ , such that ${{x}^{2}}+ax\le y$ . Prove that ${{y}^{2}}+by\ge abx$ .

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#6 h an hour ago

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Let $a,b$ be positive real numbers , such that $\frac{a+b}{2}+\frac{2ab}{a+b}+a\le b+\sqrt{ab}$ . Prove that $\frac{a+b}{2}+\frac{2ab}{a+b}+b\ge a+\sqrt{ab}$ .

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