If $\varepsilon > a + b$, then $\varepsilon^2 > (\sqrt{a}+\sqrt{b})^2$ For $a,b>0$ and $\varepsilon <1$, if $\varepsilon > a + b$, then does it follow: $\varepsilon^2 > (\sqrt{a}+\sqrt{b})^2$?
My attempt:
$(\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{ab}$ and further we obtain: $\varepsilon^{2}>a^2+2ab +b^2$
My problem is that attempting to use $\sqrt{ab}\leq ab$ does not hold if $0<ab<1$. Is the inequality just simply not true, or am I not seeing something?
 A: The inequality is simply not true. Take $a=b=0.1$ and $\epsilon = 0.3.$ Then $\epsilon^2 = 0.09$ and $(\sqrt{a}+\sqrt{b})^2 = 0.4.$
A: Its not true let $a=b=1/4$ and let $\epsilon=0.6$ then the LHS is true since $0.6>1/2$ but the RHS is false since $1>(0.6)^2$.
A: Note that the dimensionality is mismatched here. The constraint holds $\epsilon$ comparable to a linear function of $a,b$, and the lower bound is $0$, which means we can freely scale up/down a test case by multiplying everything by a large/small positive number.
In the proposed inequality $\epsilon^2 > (\sqrt a + \sqrt b)^2$, the LHS has degree two and the RHS only has degree one.  It means if we pick everything to be very small then the inequality will fail, agreeing with what you observed about $\sqrt{ab}$ for small $ab$.
Conversely, if the LHS had a lower degree than the RHS, it would be easy to make the inequality fall apart by picking very large numbers instead.  So we can expect inequalities on homogeneous constraints to be generally false unless they’re scale-invariant (or at least relatively insensitive to scalar multiplication).
