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#1 h Jun 4, 2017, 10:18 AM • 2 Y

Y by Adventure10, Mango247

For any triangle $ABC$ with angles in $(0,\pi)$ prove that:

$\cos{A}+ 2\cos{B}\cos{C} \le 1 \ \ ;$

Greetings from Lorian Saceanu

04-June 2017

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DerJan

#2 h Jun 4, 2017, 10:32 AM • 2 Y

Y by Adventure10, Mango247

Lonesan wrote:

For any triangle $ABC$ with angles in $(0,\pi)$ prove that:

$\cos{A}+ 2\cos{B}\cos{C} \le 1 \ \ ;$

Greetings from Lorian Saceanu

04-June 2017

Nice inequality! Note that
$\cos A = \cos(\pi -B-C) = -\cos(B+C) = -\cos B\cos C+\sin B\sin C.$ Thus,
$LHS = \cos B\cos C+\sin B\sin C = \cos(B-C) \leqslant 1.$

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#3 h Jun 4, 2017, 12:30 PM • 1 Y

Y by Adventure10

Thank you! Greetings!

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sqing

#4 h Jun 4, 2017, 12:51 PM • 2 Y

Y by Adventure10, Mango247

Lonesan wrote:

For any triangle $ABC$ with angles in $(0,\pi)$ prove that:

$\cos{A}+ 2\cos{B}\cos{C} \le 1 \ \ ;$

Greetings from Lorian Saceanu

04-June 2017

For any triangle $ABC$ ,prove that:

$sinA- 2sinBcosC\le 1$

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Dr Sonnhard Graubner

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Dr Sonnhard Graubner

#6 h Jun 18, 2017, 4:09 AM • 2 Y

Y by Adventure10, Mango247

hello, it is equivalent to $(a^2b^2-b^4+2b^2c^2-c^4)^2\geq 0$ Sonnhard.

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#7 h Jun 18, 2017, 5:19 AM • 1 Y

Y by Adventure10

sqing wrote:

Lonesan wrote:

For any triangle $ABC$ with angles in $(0,\pi)$ prove that:

$\cos{A}+ 2\cos{B}\cos{C} \le 1 \ \ ;$

Greetings from Lorian Saceanu

04-June 2017

For any triangle $ABC$ ,prove that:

$sinA- 2sinBcosC\le 1$

$SinA=Sin (B+C)$

$SinA-2SinBCosC=Sin (B+C)-2SinBCosC =SinCcosB-SinBCosC=Sin (C-B)\leq 1$

This post has been edited 1 time. Last edited by Weakinmath, Jun 18, 2017, 5:20 AM

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sqing

#8 h Apr 23, 2020, 1:29 AM

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$\cos{A}+ 2\cos{B}\cos{C} \le 1$ $\iff$ $\sin\frac{A}{2}+ 2\sin\frac{B}{2}\sin\frac{C}{2} \le 1$ $\iff$ $\big|1-2\sin\frac{A}{2}\big|\leq\sqrt{1-\frac{2r}{R}}.$ SXTX,Q314

$\sin\frac{A}{2}+ 2\sin\frac{B}{2}\sin\frac{C}{2} =\cos\frac{B-C}{2}$ $\frac{r}{R}=4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$

This post has been edited 1 time. Last edited by sqing, Apr 23, 2020, 1:34 AM

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#9 h Apr 25, 2020, 10:47 AM

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Very good, dear professor Qing!
Best regards!

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sqing

#10 h Apr 26, 2020, 1:40 AM • 3 Y

Y by Mango247, Mango247, Mango247

Lonesan wrote:

Very good, dear professor Qing!
Best regards!

Thank you very much.
Best regards!

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sqing

#11 h Apr 26, 2020, 8:08 AM

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Prove that in an acute triangle with angles $A, B, C$ , we have
$\sqrt[4]{\cot A}+\sqrt[4]{\cot B}+\sqrt[4]{\cot C}\geq 2.$

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42132 posts

sqing

#12 h Feb 18, 2021, 6:35 AM

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For an acute-angled $\triangle ABC$ ,we have $\sqrt{cot A}+\sqrt{cot B}+\sqrt{cot C} \ge 2.$ Prove that in an acute triangle with angles $A, B, C$ , we have
$\sqrt{cot A}+\sqrt{cot B}+\sqrt{cot C} \le \frac{\sqrt S}{r}.$ For an arbitrary triangle $ABC$ is true: $\sqrt{\cot A+ \cot B} + \sqrt{\cot B + \cot C} + \sqrt{\cot C + \cot A} \ge \sqrt{ 6 \sqrt{3} }$ h h h

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sqing

#13 h Mar 11, 2021, 3:06 PM

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sqing wrote:

For an arbitrary triangle $ABC$ is true: $\sqrt{\cot A+ \cot B} + \sqrt{\cot B + \cot C} + \sqrt{\cot C + \cot A} \ge \sqrt{ 6 \sqrt{3} }$

https://math.stackexchange.com/questions/1616462/prove-that-sum-limits-cyc-sqrt-cota-cotb-ge-2-sqrt2?rq=1

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