For any nonnegative reals $x, y$ show the inequality $$x^2y^2 + x^2y + xy^2 \le x^4y + x + y^4$$.
Problem
Source: Kyiv City MO 2022 Round 1, Problem 10.4
Tags: inequalities
24.01.2022 04:12
For any nonnegative reals $x, y$ show the inequality $$\frac{1}{2}xy(x^2+y^2 )+ x^2y + xy^2 \leq x^4y + x + y^4$$ MS_Kekas wrote: For any nonnegative reals $x, y$ show the inequality $$x^2y^2 + x^2y + xy^2 \le x^4y + x + y^4$$. Nobody solved this inequality? Why?
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24.01.2022 05:23
$(6x^4y+2x+5y^4)+(5x^4y+6x+2y^4)+(2x^4y+5x+6y^4)\ge 13(x^2y^2+x^2y+xy^2)$
24.01.2022 06:13
Adding a dummy third variable makes the inequality look less weird: $x^4y+y^4z+z^4x \ge x^2y^2z + x^2yz^2 + xy^2z^2$. The rest of the proof is a simple application of Cauchy with weights $(\frac{6}{13}, \frac{5}{13}, \frac{2}{13})$ cyclically to all three terms.
30.01.2022 13:39
For any nonnegative reals $x, y$ show the inequality $$x^2y^2 + x^2y + xy^2 \le x^3y^2 + x^2 + y^3$$
05.05.2023 15:42
MS_Kekas wrote: For any nonnegative reals $x, y$ show the inequality $$x^2y^2 + x^2y + xy^2 \le x^4y + x + y^4$$. $$x^4y + x + y^4-(x^2y^2 + x^2y + xy^2)=x(y+1)(y-1)^2+y(x+y)(x-y)^2+xy(x+1)(x-1)^2\geq 0$$p/8393267451
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