Let $a\in \mathbb [0,1]$. I want to know if the following is true $$f(y)=\inf_{y\in[0,1]}\{y-y\ln(y)\ge a\}\ne a^2.$$
So plugging in $a^2$ and rearranging yields $$a\ge \frac{1}{1-2\ln(a)}.$$ Thus there is some $a$ such that equality holds, i.e. $a^2$ is indeed the infimum. Therefore the statement is not true. Is this reasoning correct?
Actually equality does hold when $a=1$, so I will assume that $ a<1$.
You have to provide more details. $y -y\ln y $ is an increasing function on $[0,1]$ (since its derivative is non-negative). It is $0$ at $y=0$ and $1$ at $y=1$. So there is a unique point where it attains the value $a$. This point is the infimum $f$ and if $f=a^{2}$ for this value of $y$ then we must have $y-y\ln y =a$ when $y=a^{2}$. But $a^{2}-a^{2} \ln a^{2}=(1-2\ln a)a^{2}\neq a$ : I will let you verify that $x-2x \ln x$ is increasing in $(0,e^{-1/2})$ , decreasing in $(e^{-1/2},1)$ and has the values $0, 2e^{-1/2}, 1$ at the points $0, e^{-1/2}$ and $1$ respectively. This makes $ (1-2\ln a)a^{2}> a$ except when $a=1$. I will leave the case $a=1$ to you.