Prove which term is bigger in magnitude

Question

I am struggling to prove which of these two terms is bigger in magnitude.

Assumption: \begin{align} A < \theta \end{align}

Term 1: \begin{align} \sqrt{\delta^2 \theta^4 + 4 \theta^2 A (\epsilon - 1 ) (1 + \varphi)} - \sqrt{\delta^2 \theta^2 A^2 + 4 \theta A^2 (\epsilon - 1 ) (1 + \varphi)} \end{align}

Term 2: \begin{align} \delta \theta A - \delta \theta^2 \end{align}

Signs of the coefficients: All coefficients are positive and $\epsilon > 1$.

My attempt: I have noticed that if we ignore the second term under the square roots Term 1 is the negative of Term 2. So I have tried to work with inequalities but I was unsuccessful. Does anyone have any ideas how to tackle this?

Answer 1

Let $k = \frac{4}{\delta^2}(\epsilon-1)(1+\varphi)$. Now we divide both the equations by $\delta\theta$, and see that it's equivalent to

$$\sqrt{\theta^2+kA} - \sqrt{A^2+k\frac{A^2}{\theta}} \qquad\text{vs} \qquad A - \theta$$.

As @Fotis said the RHS is negative here and the LHS isn't. I'm going to assume that the question meant the absolute value of these terms, which makes the right hand side $\theta - A > 0$.

When $k=0$ we have that the LHS and RHS are equal, so all we need to show is that the LHS is increasing or decreasing with respect to $k$ and we are done.

Differentiating with respect to $k$ we get

$$\partial_k(LHS) = \frac{1}{2}\left[A(\theta^2 + kA)^{-1/2} - \frac{A^2}{\theta}(A^2 + k\frac{A^2}{\theta})^{-1/2}\right]$$ So now we want to find when

$$\theta(\theta^2 + kA)^{-1/2} > (1 + \frac{k}{\theta})^{-1/2}$$ where we have multiplied by $\frac{\theta}{A}>0$.

This happens when $$\theta^2+k\theta>\theta^2+kA$$ so... all the time!

This shows that the derivative with respect to $k$ is positive and thus the LHS increases as $k$ increases so the LHS is bigger than the RHS for all $k>0$.

Answer 2

Let $q=4θΑ(\epsilon-1)(1+φ)$
Then the first term becomes: $$X=\sqrt{δ^{2}θ^{4}+θq}-\sqrt{δ^{2}θ^{2}Α^{2}+Αq}$$ Using the fact that θ>Α we get: $$X>\sqrt{δ^{2}θ^{2}Α^{2}+Αq}-\sqrt{δ^{2}θ^{2}Α^{2}+Αq}=0 $$ $$\Rightarrow X>0$$

On the other hand, for term 2: $$Y=δθΑ-δθ^{2}$$ we have that: $$Y<δθ^2-δθ^{2}=0$$ $$\Rightarrow Y<0$$ So in the end we have: $$X>0>Y$$ Which means that the first term is bigger than the second one.