Can You Prove this Trigonometric Inequality Challenge?

[anemone]

Let the real $x\in \left(0,\,\dfrac{\pi}{2}\right)$, prove that $\dfrac{\sin^3 x}{5}+\dfrac{\cos^3 x}{12}≥ \dfrac{1}{13}$.

 

[MarkFL]

My solution:

Spoiler

Let's multiply through by:

\(\displaystyle 5\cdot12\cdot13=780\)

to get:

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

Let:

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

and:

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

Using Lagrange multipliers, we obtain:

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

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

From this, we obtain:

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

And, using the constraint, this implies:

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

Hence:

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

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

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

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

Thus, we may state:

\(\displaystyle f_{\min}=60\)

 

[anemone]

My solution:

Spoiler

Let's multiply through by:

\(\displaystyle 5\cdot12\cdot13=780\)

to get:

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

Let:

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

and:

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

Using Lagrange multipliers, we obtain:

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

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

From this, we obtain:

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

And, using the constraint, this implies:

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

Hence:

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

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

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

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

Thus, we may state:

\(\displaystyle f_{\min}=60\)

Very good job, MarkFL!(Cool)

I welcome others to try it with another approach as well!

 

[Opalg]

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]

 

[anemone]

[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:

Spoiler

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:

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