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For $x \in \left(0, \dfrac{\pi}{2}\right)$, is the minimum value of $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x} = 1$? So considering ($\dfrac{1}{\cos x}$, $\dfrac{1}{\sin x}$) and ($\sin^3x$, $\cos^3x$), is it right to use the rearrangement inequality and conclude that $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x}$ is more than or equal to $\dfrac{\sin^3x}{\sin x} +\dfrac{\cos^3x}{\cos x}$ which is equal to $1$? Thanks.

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  • $\begingroup$ @ChristianF Note that $x\in(0,\pi/2)$. $\endgroup$ – gimusi May 29 '18 at 5:38
  • $\begingroup$ Alright thanks for all the answers below! $\endgroup$ – shgdh fgxbcv May 29 '18 at 7:23
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Yes it is fine, indeed for $x\in(0,\pi/2)$ we have in both cases

  1. $\sin x\ge \cos x \implies \sin^3 x\ge \cos^3 x \implies \frac1{\sin x}\le \frac1{\cos x}$

then

$$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}\ge \frac{\sin^3 x}{\sin x}+\frac{\cos^3 x}{\cos x}=1$$

  1. $\cos x\ge \sin x \implies \cos^3 x\ge \sin^3 x \implies \frac1{\cos x}\le \frac1{\sin x}$

then

$$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}\ge \frac{\sin^3 x}{\sin x}+\frac{\cos^3 x}{\cos x}=1$$

and equality holds for $\sin x=\cos x$ that is $x=\pi/4$.

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  • $\begingroup$ Gimusi.Be kind enough to check my answer.Thanks, Peter $\endgroup$ – Peter Szilas May 29 '18 at 7:34
  • $\begingroup$ Gimusi.Thanks a lot:)) $\endgroup$ – Peter Szilas May 29 '18 at 7:38
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Your proof is right.

Also, by C-S we obtain: $$\frac{\sin^3x}{\cos x} +\frac{\cos^3x}{\sin x}=\frac{\sin^4x}{\sin{x}\cos x} +\frac{\cos^4x}{\sin x\cos{x}}\geq\frac{(\sin^2x+\cos^2x)^2}{2\sin{x}\cos{x}}=\frac{1}{\sin2x}\geq1.$$ The equality occurs for $x=\frac{\pi}{4},$ which says that $1$ is a minimal value.

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Put $a = \sin x, b = \cos x \implies a^2+b^2=1\implies \dfrac{a^2}{b}+\dfrac{b^2}{a} > \dfrac{a^2}{1} + \dfrac{b^2}{1} = a^2+b^2 = 1$ . In fact this is a different inequality, but still looks nice though. To return to the question above, observe that $\dfrac{a}{b}+\dfrac{b}{a} \ge 2$ by AM-GM inequality. Thus $(a^2+b^2)\left(\dfrac{a}{b}+\dfrac{b}{a}\right) \ge 2\implies \dfrac{a^3}{b} + \dfrac{b^3}{a} + 2ab \ge 2\implies \dfrac{a^3}{b}+\dfrac{b^3}{a} \ge 2 -2ab \ge 2 - (a^2+b^2) = 2-1 = 1$ by AM-GM inequality again.

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  • $\begingroup$ DeepSea.Nice answer! $\endgroup$ – Peter Szilas May 30 '18 at 6:23
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Correct me if wrong:

Another option.

Let $x \in (0,π/2)$.

$f(x):=$

$2 \dfrac{\sin^4 x}{2\sin x \cos x} +2 \dfrac{\cos^4 x}{2 \sin x \cos} =$

$\dfrac{2}{\sin 2x}×$

$((\sin^2x+\cos^2x)^2 -(1/2)\sin^2 2x)$

$=\dfrac{2}{\sin 2x}(1-(1/2)\sin^2 2x)$;

Let $y:= \sin 2x$, then $0 <y \le 1,$

and consider $g(y): = 2(1/y - (1/2)y)$.

Need to find the minimum of $g(y)$.

$g'(y) = 2(-1/y^2 -1/2) <0.$

$g$ is strictly decreasing

$\min_{0<y \le1} g(y)= 1.$

The minimum of the given expression is at $y=\sin 2x =1$,

i.e. at $x=π/4$.

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  • $\begingroup$ Yes it is also fine! $\endgroup$ – gimusi May 29 '18 at 7:36
  • $\begingroup$ @Peter: I think your answer is wrong. You have proven that $f(x) \ge \sin (2x)$ does not imply that $f(x) \ge 1$ because $\sin (2x) \le 1$. $\endgroup$ – DeepSea May 30 '18 at 4:52
  • $\begingroup$ DeepSea. Firstly thanks for your comment . Thought about it when posting. You are right. Reaoning is flawed. $\endgroup$ – Peter Szilas May 30 '18 at 6:18
  • $\begingroup$ DeapSea.Hopefully correct now.Thanks! $\endgroup$ – Peter Szilas May 30 '18 at 8:56
  • $\begingroup$ @PeterSzilas: It is good now.... $\endgroup$ – DeepSea May 30 '18 at 9:18
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Cauchy Schwarz Inequality

$$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}=\frac{\sin^4 x}{\cos x\sin x}+\frac{\cos^4 x}{\sin x\cos x}\geq \frac{(\sin^2 x+\cos^2 x)^2}{2\sin x\cos x}$$

$$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}\geq \frac{1}{\sin 2x}\geq 1$$

equality hold when $$\frac{\sin^2 x}{\sin x\cos x}=\frac{\cos^2 x}{\sin x\cos x}$$

that is $\displaystyle x=\frac{\pi}{4}.$

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