prove the following statement $|a-b| \le 2\sqrt{a}c + d$ implies $a \le \varepsilon a/(1+\varepsilon) + (1+\varepsilon)c^2/\varepsilon + (b + d)$

Question

I encountered a statement in a paper and I am not sure why this is the case.

It says: let $a$, $b$, $c$, $d$ be positive real numbers and $\varepsilon \in (0, 1]$, then $|a-b| \le 2\sqrt{a}c + d$ implies $a \le \varepsilon a/(1+\varepsilon) + (1+\varepsilon)c^2/\varepsilon + (b + d)$.

I tried to search online but couldn't find anything. Any reference, hint, proof etc. will be much appreciated. Thanks.

Answer 1

First note Young's inequality,

$$ \alpha\beta \leq \delta \alpha^2 + \frac{\beta^2}{4\delta}$$

For all $\alpha,\beta,\delta >0$. Then, by the triangle inequality and the above applied to $\alpha = \sqrt{a}$, $\beta = 2c$, and $\delta = \frac{\epsilon}{1+\epsilon}$,

$$a\leq |a-b|+b \leq 2c\sqrt{a}+d+b \leq \frac{\epsilon a}{1+\epsilon}+\frac{(1+\epsilon)c^2}{\epsilon}+(b+d) $$

Answer 2

Let $e=a-b-d.$ By hypothesis, $e\le2c\sqrt a,$ and we want to prove that $$e\le\varepsilon a/(1+\varepsilon) + (1+\varepsilon)c^2/\varepsilon.$$ It is sufficient to prove that $$2c\sqrt a\le\varepsilon a/(1+\varepsilon) + (1+\varepsilon)c^2/\varepsilon.$$ For this, simply notice that $$\varepsilon a/(1+\varepsilon)-2c\sqrt a+ (1+\varepsilon)c^2/\varepsilon=\left(\sqrt\frac{a\varepsilon}{1+\varepsilon}-c\sqrt\frac{1+\varepsilon}\varepsilon\right)^2.$$