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anemone
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Let the real $x\in \left(0,\,\dfrac{\pi}{2}\right)$, prove that $\dfrac{\sin^3 x}{5}+\dfrac{\cos^3 x}{12}≥ \dfrac{1}{13}$.
MarkFL said:My solution:
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:
\(\displaystyle f_{\min}=60\)
[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 said:Let the real $x\in \left(0,\,\dfrac{\pi}{2}\right)$, prove that $\dfrac{\sin^3 x}{5}+\dfrac{\cos^3 x}{12}≥ \dfrac{1}{13}$.
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]
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