IRL_ABooklet

$$f(t)=\frac{\sqrt{1+t}-\sqrt{|1-t|}}{1+\sqrt{t}}$$ and observe that $f(1/t)=f(t)$, and that $f(0)=0,f(1)=1/\sqrt{2}$. Therefore, it suffices to prove that $2(f(s^2){)}^2\le 1$ if $0\le s\le 1$, i.e, that $$2(1+s^2+1-s^2-2\sqrt{1-s^4})\le (1+s{)}^2.$$ This is equivalent to $4-(1+s{)}^2\le 4\sqrt{1-s^4}$, or $(4-(1+s{)}^2{)}^2\le 16(1-s^4)$, i.e. $$(1-s{)}^2(3+s{)}^2\le 16(1-s)(1+s+s^2+s^3).$$ Since $0\le s\le 1$ this holds provided that $$(1-s)(9+6s+s^2)\le 16(1+s+s^2+s^3),$$ equivalently, iff $$9-3s-5s^2-s^3\le 16(1+s+s^2+s^3),$$ which clearly holds for all $s\ge 0$.

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Alternative solution.

We first show that $$\sqrt{a}-\sqrt{b}\le \sqrt{a-b}\text{if}0\le b\le a.(9)$$ Indeed, we have $0\le \sqrt{b}\le \sqrt{a}$ and multiply by $2\sqrt{b}$ to get $0\le 2b\le 2\sqrt{ab}$. If we add $a-b-2\sqrt{ab}$ to both sides this yields $$(\sqrt{a}-\sqrt{b}{)}^2=a-2\sqrt{ab}+b\le a-b.$$ Because $a\ge b$, we can take square roots to obtain $\sqrt{a}-\sqrt{b}\le \sqrt{a-b}$. Next, we note that the inequality we wish to show is unchanged when $t$ is replaced by $1/t$. Hence, it suffices to consider $0\le t\le 1$. In this case, $|1-t|=1-t$. With $a=1+t$ and $b=1-t$ we obtain from (9) the inequality $$\sqrt{1+t}-\sqrt{1-t}\le \sqrt{2t}=\frac{2\sqrt{t}}{\sqrt{2}}\le \frac{1+\sqrt{t}}{\sqrt{2}},$$ where we have used $\sqrt{t}\le 1$ in the last step. This is the desired result.