Given $y(0)=y(1)=0,~|y'(x)|<1$ for $0\le x\le 1$ is it true that $\int_0^1\frac1{(1+y')^2}dx\ge 1$?

Question

Q1. Given $y(0)=y(1)=0,~|y'(x)|<1$ for $0\le x\le 1$ is it true that $\int_0^1\frac1{(1+y')^2}dx\ge 1$?

I checked this for 2 functions

$y(x)=\frac{a}\pi \sin\pi x,~|a|<1,~\int_0^1\frac1{(1+y')^2}dx=\frac1{(1-a^2)^{3/2}}$,

$y(x)=a x(1-x),~|a|<1,~\int_0^1\frac1{(1+y')^2}dx=\frac1{1-a^2}$,

but couldn't find any counterexamples or the proof.

I'm also interested in this question

Q2. Given $y(0)=y(1)=0,~y'(x)>-1$ for $0\le x\le 1$ is it true that $\int_0^1\frac1{(1+y')^2}dx\ge 1$?

Answer 1

A1. The function $\varphi(s) := \dfrac{1}{(1+s)^2}$ is convex in $(-1, +\infty)$. By Jensen's inequality we have that $$ \int_0^1 \varphi(y'(x))\, dx \geq \varphi\left(\int_0^1 y'(x)\, dx\right) = \varphi(0) = 1. $$

A2. See A1.