Proving an inequality involving cosine functions

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Proving an inequality involving cosine functions

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#1 h Jan 22, 2025, 7:15 PM

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Hello AoPS Community,

I am curious about how to demonstrate the following inequality:

\sqrt{1 - |\cos(x \pm y)|^a} \leq \sqrt{1 - |\cos(x)|^a} + \sqrt{1 - |\cos(y)|^a}, \quad \text{for } a \geq 1.

I’ve plotted the functions, and the inequality seems to hold. However, I am looking for a rigorous way to prove it.

I’d greatly appreciate insights into how to break this down analytically and references to similar problems or techniques that might be helpful.

Thank you so much for your guidance and support!

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#2 h Jan 23, 2025, 3:24 PM

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When I started writing this reply I thought I had a proof for the inequality, but there was a mistake. So probably the following does not help much. Nevertheless let me write down how I would have started an argument:

We want to show that
$\sqrt{1 - |\cos(x \pm y)|^a} \leq \sqrt{1 - |\cos(x)|^a} + \sqrt{1 - |\cos(y)|^a}, \quad \text{for all } a \ge 1, x,y \in R,$ i.e.
$\sqrt{1 - |st \pm \sqrt{1-s^2} \sqrt{1-t^2}|^a} \leq \sqrt{1 - |s|^a} + \sqrt{1 - |t|^a}, \quad \text{for all } a \ge 1, s,t \in [-1,1],$ w.l.o.g $s,t \ge 0$ and $\pm$ ist $-$ on the LHS, i.e. we need to show
$\sqrt{1 - |st - \sqrt{1-s^2} \sqrt{1-t^2}|^a} \leq \sqrt{1 - s^a} + \sqrt{1 - t^a}, \quad \text{for all } a \ge 1, s,t \in [0,1],$ For further simplification we fix $c \in [-1,1]$ and only consider $s,t\in [0,1]$ with $st - \sqrt{1-s^2} \sqrt{1-t^2} = c$ , which is equivalent to $st \ge c$ and $g(s,t) := s^2+t^2-2cst + c^2- 1 = 0$ . With respect to this constraint we want to minize the RHS $f(s,t) = \sqrt{1 - s^a} + \sqrt{1 - t^a}$ . This can be done via Lagrange. $f$ is continuous on $D = \{(s,t) \in [0,1]^2: st \ge c\}$ and the constraint $M = \{(s,t) \in D: g(s,t) = 0\}$ defines a compact set, so $f$ attains a minimum on $M$ .

Lagrange gives that if the minimum is attained at $(s,t) \in M$ , then $(s,t) \in \partial D$ or $\nabla g(s,t) = 0$ or $\nabla f(s,t) = \lambda \nabla g(s,t)$ for some real $\lambda$ .

For $(s,t) \in \partial D$ we have $s = 0, t = 0, s = 1, t = 1$ and $st = c$ (and the latter implies $s = 1$ or $t = 1$ ), and in all of these cases it is easy to check, that the inequality holds.
For the other conditions we compute
$\nabla g(s,t) = (2s-2ct, 2t - 2cs), \nabla f(s,t) = (\frac{-as^{a-1}}{2 \sqrt{1-s^a}}, \frac{-at^{a-1}}{2 \sqrt{1-t^a}})$ In case of $\nabla g(s,t) = 0$ we thus get $s = ct$ and $t = cs$ , i.e. $s = c^2s$ and thus $s = 0$ or $c = \pm 1$ . The latter implies that $s = t = 1$ or $s = t = 0$ , so the inequality holds by the above.

In case of $\nabla f(s,t) = \lambda \nabla g(s,t)$ I attempted to show that $s = t$ . But this was nonsense, and this makes the approach probably futile.

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#3 h Jan 24, 2025, 11:54 AM

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Here is a new and hopefully more successful attempt: As shown above we need to show that
$\sqrt{1 - |st - \sqrt{1-s^2} \sqrt{1-t^2}|^a} \leq \sqrt{1 - s^a} + \sqrt{1 - t^a}, \quad \text{for all } a \ge 1, s,t \in [0,1].$ Note that $1-s^a$ is decreasing in $s$ and $st- \sqrt{1-s^2} \sqrt{1-t^2}$ is increasing in $s$ . So for $st- \sqrt{1-s^2} \sqrt{1-t^2}< 0$ it suffices to consider the worst case
$st- \sqrt{1-s^2} \sqrt{1-t^2}=0$ . Thus we may assume $st- \sqrt{1-s^2} \sqrt{1-t^2} \ge 0$ , i.e. $s^2+t^2 \ge 1$ . Let
$f(s,t) = \sqrt{1 - s^a} + \sqrt{1 - t^a} - \sqrt{1 - (st - \sqrt{1-s^2} \sqrt{1-t^2})^a} \text{ and } D = \{(s,t) \in [0,1]^2 : s^2+t^2 \ge 1\}.$ We need to show that $f(s,t) \ge 0$ for $(s,t) \in D$ . This can be done by brute force using simple multivariable calculus.

First note that by continuity and compactness $f$ attains its minimum on $D$ . Also note that $f \ge 0$ on the boundary: If $s = 1$ or $t = 1$ we have equality and
for $s^2 + t^2 = 1$ we need to check that $\sqrt{1-s^a} + \sqrt{1-t^a} \ge 1$ . By monotonicity the worst case is $a = 1$ , but in this case the inequality is equivalent to $2\sqrt{1-s^a} \sqrt{1-t^a} \ge 1-s-t$ , which clearly holds since $s+t \ge s^2 + t^2 = 1$ .

Thus it remains to theck the inequality for critical points in the interior of $D$ . We get
$0 = \partial_1 f(s,t) = \frac{a(st - \sqrt{1-s^2} \sqrt{1-t^2})^{a-1}} {2\sqrt{1 - (st - \sqrt{1-s^2} \sqrt{1-t^2})^a} } (t +s \frac{\sqrt{1-t^2}}{ \sqrt{1-s^2}})- \frac{as^{a-1}}{2 \sqrt{1-s^a}}$ and similar for $\partial_2 f(s,t)$ . This may look horrible, but the most horrible parts are symmetric in $s,t$ . In particular the above implies that
$\frac{as^{a-1} \sqrt{1-s^2}}{2 \sqrt{1-s^a}} = \frac{at^{a-1} \sqrt{1-t^2}}{2 \sqrt{1-t^a}}.$ We want to show that this implies $s = t$ . For this we check that $g(x) = \frac{x^{2a-2}(1-x^2)}{(1-x^a)}$ is strictly increasing in $x \in (0,1)$ . Using the logarithmic derivative it suffices to show that
$\frac{2a-2}x - \frac{2x}{1-x^2} + \frac {ax^{a-1}}{1-x^a} > 0 \Leftrightarrow 2a-2 - \frac{2x^2}{1-x^2} + \frac {ax^{a}}{1-x^a} > 0 \Leftrightarrow a - \frac{2}{1-x^2} + \frac {a}{1-x^a} > 0$ and we note that for $a \in [1,2]$ we have $a - \frac{2}{1-x^2} + \frac {a}{1-x^a} \ge 1 - \frac{2}{1-x^2} + \frac {1}{1-x} = \frac{x-x^2}{1-x^2}>0$ . And we check that for $a \ge 2$ we have
$\frac {a}{1-x^a} \ge \frac {2}{1-x^2} \Leftrightarrow a-2 \ge ax^2-2x^a \Leftrightarrow a-2 \ge x^2(a - 2x^{a-2}),$ which is obvious.

It now remains to show that $f(s,t) \ge 0$ for all $(s,t) \in D$ with $s = t$ , i.e. $f(t,t) \ge 0$ for $t \in [\frac 1 {\sqrt 2},1]$ . This is equivalent to
$2 \sqrt{1-t^a} \ge \sqrt{1 - (2t^2-1)^a} \Leftrightarrow 4-4t^a \ge 1-(2t^2-1)^a \Leftrightarrow \frac 3 4 1^a+ \frac 1 4 (2t^2-1)^a \ge t^a$ This follows from convexity and monotonicity of the function $x^a$ since $\frac 3 4 + \frac 1 4 (2t^2-1) \ge t$ , which is easy to see.

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#4 h Yesterday at 3:33 PM

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Thank you so much for your detailed and thoughtful response! Apologies for the delay in getting back, but your explanation was very helpful and has clarified the approach for me. I really appreciate the time and effort you put into breaking it down! It was smart

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