TITLE: Condition for trigonometric inequality
QUESTION [7 upvotes]: I want to prove the following statement:
Suppose $\frac{1}{4}(\cos(\theta_1)+\cos(\theta_2))^2+\lambda^2(a\sin(\theta_1)+b\sin(\theta_2))^2\leq 1$ holds for all $\theta_1,\theta_2\in[-\pi,\pi]$, then we should have $\lambda^2(a^2+b^2)\leq \frac{1}{2}$.
This problem appears when I want to find the stability condition for a numerical scheme. I tried to use Lagrange multiplier, but it turns out to be very complicated. I have also tried to find some specific $\theta_1,\theta_2$, so that the first inequality can imply the second, but I failed to do so.

REPLY [2 votes]: Here is an answer which shows the OP's claim.
Say A is the first condition and B is the second. The claim is that there is an implication $A \to B$. This is logically equivalent to the implication Not B $\to$ Not A. So  one might show that one. 
In full prose, the equivalent condition reads:
Suppose $\lambda^2(a^2+b^2) > \frac{1}{2}$. Then there exists some $\theta_1,\theta_2\in[-\pi,\pi]$ such that 
$\frac{1}{4}(\cos(\theta_1)+\cos(\theta_2))^2+\lambda^2(a\sin(\theta_1)+b\sin(\theta_2))^2 > 1$.
Let us instead write the assumption as $\lambda^2(a^2+b^2) \geq R > \frac{1}{2}$. The extra $R$ has been put in to show that indeed, for our argument,  $R>1/2$ is necessary, and no smaller $R$ will be sufficient.
Proof:
Since, by assumption, 
$\lambda^2 \geq R / (a^2+b^2)$, we can show
$$\frac{1}{4}(\cos(\theta_1)+\cos(\theta_2))^2+R \frac{(a\sin(\theta_1)+b\sin(\theta_2))^2}{(a^2+b^2)} - 1> 0$$
Setting $b = q\, a$, this becomes
$$\frac{1}{4}(\cos(\theta_1)+\cos(\theta_2))^2+R \frac{(\sin(\theta_1)+q\sin(\theta_2))^2}{(1+q^2)} - 1> 0$$
Let us now set $\theta_2 = q \theta_1$. We will show that for this choice, the inequality can always be observed, given $R >1/2$. We obtain 
$$\frac{1}{4}(\cos(\theta_1)+\cos(q \theta_1))^2+R \frac{(\sin(\theta_1)+q\sin(q \theta_1))^2 }{1+q^2} -1 > 0$$
Clearly, for $\theta_1 = 0$, the LHS $= 0$. Let us therefore choose a very small $\theta_1$.  By a Taylor expansion of the LHS about $\theta_1 = 0$, we get 
LHS = $(1+q^2)\, (R-1/2) \, \theta_1^2 \;  + \cal O (\theta_1^4)$
So clearly, we have the above condition $R > 1/2$. 
If this condition is observed  we have that, for small enough $\theta_1$, the LHS $> 0$. 
This proves the OP's claim. Moreover, it is shown that with the given line of argument, the $R$ in the bound cannot be made smaller.