Carefully Constructed Inequality

Relevant wiki: Completing the Square - Intermediate

$x^2+2y^2+\dfrac{1}{2}\leq x(2y+1)$

Multiply both sides By 2

$2x^2 +4y^2 +1 \leq 4xy +2x$

$2x^2 -4xy +4y^2 -2x +1 \leq 0$

$(x^2 -4xy +4y^2) +(x^2 -2x +1) \leq 0$

${(x -2y)}^{2} + {(x-1)}^{2} \leq 0$

The sum of 2 squares can never be negative for real numbers

Therefore the expression must equal 0

${(x -2y)}^{2} + {(x-1)}^{2} = 0$

Sum of the squares of two real numbers is zero only when the two real numbers themselves are zero

$x = 1$

$P=x^2+2y^2+\dfrac{1}{2}\leq x(2y+1) \\ \Leftrightarrow P=\dfrac{x^2}{2}-x+\dfrac{1}{2}+\dfrac{x^2}{2}-2xy+2y^2\leq 0\\ \Leftrightarrow P=\dfrac{x^2-2x+1}{2}+\dfrac{x^2-4xy+4y^2}{2}\leq 0\\ \Leftrightarrow P=\dfrac{(x-1)^2}{2}+\dfrac{(x-2y)^2}{2}\leq 0\\ \text{But P is always greater than or equal to 0} \\ \Rightarrow \left\{\begin{array}{ll} x-1=0\\ x-2y=0 \end{array}\right.\\ \Rightarrow \color{#D61F06}{x=1}$