Doubt

1) How to prove this one: 2a+b+2b+c+2c+a \frac{2}{a+b} + \frac{2}{b+c} + \frac{2}{c+a} \le 1a+1b+1c \frac{1}{a} + \frac{1}{b} + \frac{1}{c}

2) Eliminate t: x=10(tsint)x=10( t-\sin t) and y=10(1cost)y=10(1-\cos t)

Note by Baibhab Chakraborty
2 months, 3 weeks ago

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Comments

This is a equation of a cycloid . I have eliminated tt but equation is Messier. Generally , x=r(tsin(t))x=r(t-\sin(t))

,and y=r(1cos(t))    cos(t)=1yr    1sin2(t)=1yray=r(1-\cos(t))\implies \cos(t)= 1-\dfrac{y}{r} \implies \sqrt{1-\sin^2(t)} = 1-\dfrac{y}{r} \rightarrow{a}

And also t=cos1(1yr)t=\cos^{-1} (1-\dfrac{y}{r})

From xx we can find sin(t)=txr=cos1(1yr)xr\sin(t)= t-\dfrac{x}{r} = \cos^{-1}(1-\dfrac{y}{r})-\dfrac{x}{r}

Putting this on aa we will find equation is x2+y2r22r(xcos1(yr1)+y)+(cos1(yr1))2=0\dfrac{x^2+y^2}{r^2} -\frac{2}{r}(x \cos^{-1}(\dfrac{y}{r}-1)+y) +(\cos^{-1} (\dfrac{y}{r}-1))^2 =0

Dwaipayan Shikari - 2 months, 3 weeks ago

Please note that I am actually trying to derive the equation of the path traced by a point fixed on a rolling body in pure rolling in space. Please help, I am a novice.

Baibhab Chakraborty - 2 months, 3 weeks ago

Ok thank you sir so much.

Baibhab Chakraborty - 2 months, 3 weeks ago

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Thanks bro . But I am not a sir!;)

Dwaipayan Shikari - 2 months, 3 weeks ago

.

A Former Brilliant Member - 2 months, 2 weeks ago

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Is there any available facility of private chat in brilliant? Just intrigued.

Baibhab Chakraborty - 2 months, 2 weeks ago

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No, there is no such feature (tho there should be lol)

A Former Brilliant Member - 2 months, 2 weeks ago

@Percy Jackson I found a solution to the first one: I had thought of using H.M but couldn;t end doing the problem. Its as follows:

a+b22aba+b \frac{a+b}{2} \ge \frac{2ab}{a+b} [ A.M \ge H.M ]

a+b2ab2a+b \frac{a+b}{2ab} \ge \frac{2}{a+b}

1a+1b22a+b \frac{\frac{1}{a} + \frac{1}{b}}{2} \ge \frac{2}{a+b}

Σ1a+1b2Σ2a+b \Sigma \frac{\frac{1}{a}+\frac{1}{b}}{2} \ge \Sigma \frac{2}{a+b}

2a+b+2b+c+2c+a1a+1b+1c \frac{2}{a+b} + \frac{2}{b+c} + \frac{2}{c+a} \le \frac{1}{a} + \frac{1}{b} + \frac{1}{c}

Baibhab Chakraborty - 2 months, 1 week ago
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