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[1] Is there an easy way to formally prove that, $$ 2xy^{2} +2x^{2} y-2x^{2} y^{2} -4xy+x+y\ge -x^{4} -y^{4} +2x^{3} +2y^{3} -2x^{2} -2y^{2} +x+y$$ $${0<x,y<1}$$ without resorting to checking partial derivatives of the quotient formed by the two sides, and finding local maxima? [2] Similarly, is there an easy way for finding $$\max_{0<x,y<1} [f(x,y)]$$ where, $$f(x,y)=2x(1+x)+2y(1+y)-8xy-4(2xy^{2} +2x^{2} y-2x^{2} y^{2} -4xy+x+y)^{2}$$ |
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Your first question: With a little manipulation you get that it is equivalent to $$x^2((1-x)^2+1)+y^2((1-y)^2+1) \ge 2xy[(1-x)(1-y)+1].$$ This can be obtained from addition of two inequalities $$x^2(1-x)^2+y^2(1-y)^2 \ge 2xy(1-x)(1-y)$$ $$x^2+y^2\ge 2xy.$$ Both of them are special cases of $a^2+b^2\ge 2ab$, which follows from $(a-b)^2\ge 0$. (Or, if you prefer, you can consider it as a special case of AM-GM inequality.) Note: To check the algebraic manipulations, you can simply compare the results for 2xy^2 +2x^2 y-2x^2 y^2 -4xy+x+y - ( -x^4 -y^4 +2x^3 +2y^3 -2x^2 -2y^2 +x+y) expand x^2((1-x)^2+1)+y^2((1-y)^2+1) -2xy[(1-x)(1-y)+1] Or simply subtract the two expressions: I did not succeed in finding similar type of solution for your second problem. |
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Certainly, there is no need for taking the quotient, since $a \ge b \Leftrightarrow \min \{a-b\} \ge 0$. Here's a cool trick called the S.O.S. (sum of squares) method. The idea is to try and factor out $(x-y)^2$: $$\begin{align} LHS-RHS &=(x^4+y^4-2x^2y^2)-2(x^3+y^3-x^2y-xy^2)+2(x^2+y^2-2xy)\\ &=(x^2-y^2)^2-2(x^2-y^2)(x-y)+2(x-y)^2\\ &=(x-y)^2(x+y)^2-2(x-y)^2(x+y)+2(x-y)^2\\ &=(x-y)^2((x+y)^2-2(x+y)+2)\\ &=(x-y)^2((x+y-1)^2+1)\\ &\ge 0 \end{align}$$ Note that this holds for all $x, y \in \mathbb R$. |
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