MHB Can You Prove this Trigonometric Inequality Challenge?

AI Thread Summary
The discussion centers on proving the trigonometric inequality $\dfrac{\sin^3 x}{5}+\dfrac{\cos^3 x}{12}≥ \dfrac{1}{13}$ for $x \in \left(0,\,\dfrac{\pi}{2}\right)$. One participant employs the Cauchy–Schwarz inequality, transforming the inequality into a form involving vectors and their norms. By defining specific angles and using trigonometric identities, they demonstrate that the inequality holds true. The conversation encourages others to explore different approaches to the problem. Overall, the thread highlights collaborative problem-solving in trigonometry.
anemone
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Let the real $x\in \left(0,\,\dfrac{\pi}{2}\right)$, prove that $\dfrac{\sin^3 x}{5}+\dfrac{\cos^3 x}{12}≥ \dfrac{1}{13}$.
 
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My solution:

Let's multiply through by:

$$5\cdot12\cdot13=780$$

to get:

$$156\sin^3(x)+65\cos^3(x)\ge60$$

Let:

$$g(x,y)=x+y-\frac{\pi}{2}=0$$

and:

$$f(x,y)=156\sin^3(x)+65\sin^3(y)$$

Using Lagrange multipliers, we obtain:

$$468\sin^2(x)\cos(x)=\lambda$$

$$195\sin^2(y)\cos(y)=\lambda$$

From this, we obtain:

$$12\sin^2(x)\cos(x)=5\sin^2(y)\cos(y)$$

And, using the constraint, this implies:

$$\sin(x)=\frac{5}{13},\,\sin(y)=\frac{12}{13}$$

Hence:

$$f(x,y)=156\left(\frac{5}{13}\right)^3+65\left(\frac{12}{13}\right)^3=60$$

To show this is a minimum, we can pick another point on the constraint:

$$(x,y)=\left(\frac{\pi}{4},\frac{\pi}{4}\right)$$

$$f\left(\frac{\pi}{4},\frac{\pi}{4}\right)=\frac{156}{2\sqrt{2}}+\frac{65}{2\sqrt{2}}=\frac{221}{2\sqrt{2}}>60$$

Thus, we may state:

$$f_{\min}=60$$
 
MarkFL said:
My solution:

Let's multiply through by:

$$5\cdot12\cdot13=780$$

to get:

$$156\sin^3(x)+65\cos^3(x)\ge60$$

Let:

$$g(x,y)=x+y-\frac{\pi}{2}=0$$

and:

$$f(x,y)=156\sin^3(x)+65\sin^3(y)$$

Using Lagrange multipliers, we obtain:

$$468\sin^2(x)\cos(x)=\lambda$$

$$195\sin^2(y)\cos(y)=\lambda$$

From this, we obtain:

$$12\sin^2(x)\cos(x)=5\sin^2(y)\cos(y)$$

And, using the constraint, this implies:

$$\sin(x)=\frac{5}{13},\,\sin(y)=\frac{12}{13}$$

Hence:

$$f(x,y)=156\left(\frac{5}{13}\right)^3+65\left(\frac{12}{13}\right)^3=60$$

To show this is a minimum, we can pick another point on the constraint:

$$(x,y)=\left(\frac{\pi}{4},\frac{\pi}{4}\right)$$

$$f\left(\frac{\pi}{4},\frac{\pi}{4}\right)=\frac{156}{2\sqrt{2}}+\frac{65}{2\sqrt{2}}=\frac{221}{2\sqrt{2}}>60$$

Thus, we may state:

$$f_{\min}=60$$

Very good job, MarkFL!(Cool)

I welcome others to try it with another approach as well! :o
 
anemone said:
Let the real $x\in \left(0,\,\dfrac{\pi}{2}\right)$, prove that $\dfrac{\sin^3 x}{5}+\dfrac{\cos^3 x}{12}≥ \dfrac{1}{13}$.
[sp]Let $\theta = \arcsin\frac5{13}$, so that $\sin\theta = \frac5{13}$ and $\cos\theta = \frac{12}{13}.$ Then (after dividing the denominators by $13$) the inequality becomes $$\frac{\sin^3x}{\sin\theta} + \frac{\cos^3x}{\cos\theta} \geqslant 1.$$ To see that this is true, let $\mathbf{a}$, $\mathbf{b}$ be the vectors given by $$\mathbf{a} = \left(\sqrt{\frac{\sin^3x}{\sin\theta}}, \sqrt{\frac{\cos^3x}{\cos\theta}}\right), \qquad \mathbf{b} = \left(\sqrt{\sin x\sin\theta}, \sqrt{\cos x\cos\theta}\right).$$ The Cauchy–Schwarz inequality says that $(\mathbf{a\cdot b})^2 \leqslant \|\mathbf{a}\|^2\|\mathbf{b}\|^2$. But $$\mathbf{a\cdot b} = \sin^2x + \cos^2x = 1,$$ $$\|\mathbf{a}\|^2 = \frac{\sin^3x}{\sin\theta} + \frac{\cos^3x}{\cos\theta},$$ $$\|\mathbf{b}\|^2 = \sin x\sin\theta + \cos x\cos\theta = \cos(\theta - x).$$ Therefore $$1 \leqslant \left(\frac{\sin^3x}{\sin\theta} + \frac{\cos^3x}{\cos\theta}\right) \cos(\theta - x) \leqslant \frac{\sin^3x}{\sin\theta} + \frac{\cos^3x}{\cos\theta}.$$[/sp]
 
Opalg said:
[sp]Let $\theta = \arcsin\frac5{13}$, so that $\sin\theta = \frac5{13}$ and $\cos\theta = \frac{12}{13}.$ Then (after dividing the denominators by $13$) the inequality becomes $$\frac{\sin^3x}{\sin\theta} + \frac{\cos^3x}{\cos\theta} \geqslant 1.$$ To see that this is true, let $\mathbf{a}$, $\mathbf{b}$ be the vectors given by $$\mathbf{a} = \left(\sqrt{\frac{\sin^3x}{\sin\theta}}, \sqrt{\frac{\cos^3x}{\cos\theta}}\right), \qquad \mathbf{b} = \left(\sqrt{\sin x\sin\theta}, \sqrt{\cos x\cos\theta}\right).$$ The Cauchy–Schwarz inequality says that $(\mathbf{a\cdot b})^2 \leqslant \|\mathbf{a}\|^2\|\mathbf{b}\|^2$. But $$\mathbf{a\cdot b} = \sin^2x + \cos^2x = 1,$$ $$\|\mathbf{a}\|^2 = \frac{\sin^3x}{\sin\theta} + \frac{\cos^3x}{\cos\theta},$$ $$\|\mathbf{b}\|^2 = \sin x\sin\theta + \cos x\cos\theta = \cos(\theta - x).$$ Therefore $$1 \leqslant \left(\frac{\sin^3x}{\sin\theta} + \frac{\cos^3x}{\cos\theta}\right) \cos(\theta - x) \leqslant \frac{\sin^3x}{\sin\theta} + \frac{\cos^3x}{\cos\theta}.$$[/sp]

Very well done Opalg! Thanks for participating!(Cool)

My solution:
First, multiply the first and second fraction (top and bottom) from the LHS of the intended inequality by $\sin x$ and $\cos x$ respectively to get:

$\dfrac{\sin^3 x}{5}+\dfrac{\cos^3 x}{12}=\dfrac{\sin^4 x}{5\sin x}+\dfrac{\cos^4 x}{12 \cos x}$(*)

Since $\sin x$ and $\cos x$ are both positive reals in the given domain, we can apply the Titu's Lemma to the RHS of (*) to get:

$\begin{align*}\dfrac{\sin^4 x}{5\sin x}+\dfrac{\cos^4 x}{12 \cos x}&\ge \dfrac{(\sin^2 x+\cos^2 x)^2}{5\sin x+12 \cos x}\\&\ge \dfrac{1}{\sqrt{5^2+12^2}\sin \left(x+\tan^{-1}\dfrac{12}{5}\right)}\\&\ge \dfrac{1}{13 \sin \left(x+\tan^{-1}\dfrac{12}{5}\right)}\\&\ge \dfrac{1}{13}\,\,\,\text{since}\,\,\,\sin \left(x+\tan^{-1}\dfrac{12}{5}\right)\le 1 \,\,\,\text{for}\,\,\, x\in \left(0,\,\dfrac{\pi}{2}\right)\end{align*}$
 
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