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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#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

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

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#3 h Wednesday at 6:06 PM

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

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