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Prove that

$ \sin \frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2} \le \frac18 $

if $\alpha,\beta,\gamma$ are angles in triangle $(\alpha+\beta+\gamma=180^\circ)$

matematicarbih
  • 19

4 Answers4

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Using the half-angle formulas then your inequality is equivalent to $$\sqrt{\frac{(s-b)(s-c)}{bc}}\sqrt{\frac{(s-a)(s-c)}{ac}}\sqrt{\frac{(s-a)(s-b)}{ab}}\le \frac{1}{8}$$ can you finish? This is equivalent to$$8(s-a)(s-b)(s-c)\le abc$$ and this is $$(-a+b+c)(a-b+c)(a+b-c)\le abc$$ now we use the Ravi Substitution and we have $$(x+y)(x+z)(z+x)\geq 8xyz$$ now use $$AM-GM$$ and the proof is finished.

I have another proof, note that $$f(x)=\log(\sin(x))$$ is concave, since $$f''(x)=-\frac{1}{\sin^2(x)}$$ and then we get $$\log\left(\sin\left(\dfrac\alpha2\right)\right)+\log\left(\sin\left(\dfrac\beta2\right)\right)+\log\left(\sin\left(\dfrac\gamma2\right)\right)\le 3\log\left(\sin\left(\dfrac\pi6\right)\right)=\log\left(2^{-3}\right)$$ ready

Dr. Sonnhard Graubner
  • 95,283
2

By AM-GM and Jensen inequalities in succession, $$\prod_{cyc} \sin \frac{\alpha}2 \leq\left[ \frac{\sum_{cyc} \sin \frac{\alpha}{2}}{3}\right]^3 \leq \left[ \frac{ 3 \sin \frac{\alpha+\beta+\gamma}{6}}{3}\right]^3=\frac18$$

Nemo
  • 1,270
1

Hint:

Let $2y=2\sin\dfrac\alpha2\sin\dfrac\beta2\sin\dfrac\gamma2=\sin\dfrac\alpha2\left(\cos\dfrac{\beta-\gamma}2-\sin\dfrac\alpha2\right)$

$$\iff\sin^2\dfrac\alpha2-\sin\dfrac\alpha2\cos\dfrac{\beta-\gamma}2+2y=0$$ which is a Quadratic Equation in $\sin\dfrac\alpha2$

As $\sin\dfrac\alpha2$ is real, the discriminant must be $\ge0$

lab bhattacharjee
  • 274,582
  • See also : https://math.stackexchange.com/questions/952893/cos-a-cos-b-cos-c-leq-frac18 and https://math.stackexchange.com/questions/949530/in-triangle-abc-show-that-1-lt-cos-a-cos-b-cos-c-le-frac-32 and https://math.stackexchange.com/questions/639890/minimum-value-of-cosa-b-cosb-c-cosc-a-is-3-2 – lab bhattacharjee Mar 05 '18 at 16:45
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In the standard notation we need to prove that $$\prod_{cyc}\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}{2}}\leq\frac{1}{8}$$ or $$\prod_{cyc}\sqrt{\frac{(a+b-c)(a+c-b)}{bc}}\leq1$$ or $$\prod_{cyc}(a+b-c)\leq abc$$ or $$\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0,$$ which is Schur.

The inequality $$\prod_{cyc}(a+b-c)\leq abc$$ we can prove also by the Ravi's substitution.

Indeed, let $a=y+z$, $b=x+z$ and $c=x+y$.

Thus, $x$, $y$ and $z$ are positives and we need to prove that $$\prod_{cyc}(x+y)\geq8xyz,$$ which is obvious by AM-GM: $$x+y\geq2\sqrt{xy}.$$

Also, we can use the following way.

We need to prove that $$\sin\frac{\alpha}{2}\left(\cos\frac{\beta-\gamma}{2}-\sin\frac{\alpha}{2}\right)\leq\frac{1}{4}$$ or $$4\sin^2\alpha+1\geq4\sin\frac{\alpha}{2}\cos\frac{\beta-\gamma}{2},$$ which is true by AM-GM: $$4\sin^2\alpha+1\geq4\sin\frac{\alpha}{2}\geq4\sin\frac{\alpha}{2}\cos\frac{\beta-\gamma}{2}.$$ Also, the starting inequality is equivalent to $$R\geq2 r,$$ which is obvious.

Michael Rozenberg
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