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Some awesome problems that I loved to solve, Try solving them they are fun...

(\(1\))

If \(a_1,a_2,a_3...............a_{24}\) are such that

\(\dfrac{a_2}{a_1}= \dfrac{a_3}{a_2}=............ =\dfrac{a_{24}}{a_{23}}\)

and \(\displaystyle\sum_{i=1}^{24}\)\(a_i = 2\)

\(\displaystyle\sum_{i=1}^{24}\)\(\dfrac{1}{a_i} = 1\)

Find \(\displaystyle\prod_{i=1}^{24}\)\(a_i\)

(\(2\))

For an acute-angled \(\triangle{ABC}\) ] \(0,\pi/2\)] , proove that (without using AM-GM)

\(\sin(A). \sin(B). \sin(C) \leq \dfrac{3\sqrt3}{8}\)

(\(3\))

Find the largest constant \(k\) such that

\(\dfrac{kabc}{a+b+c} \leq (a+b) ^2+(a+b+4c)^2\)

(\(4\))

Find the real roots of the equation

\(x^2+2ax+\dfrac{1}{16} = - a+\sqrt{a^2+x-\dfrac{1}{16}}\)

(\(5\))

Proove that

\(1 <\dfrac{1}{1001} + \dfrac{1}{1002} +............ +\dfrac{1}{3001} < \dfrac{4}{3}\)


Hope you will enjoy them!!

Note by Parth Lohomi
2 years, 7 months ago

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Solution for( 2)

F(x)=\(\log(\sin(x))\)

F'(x)=\(\cot(x)\)

F"(x)= -\(cosec(x)\)

Apply jensens inequality

\(F(\dfrac{A+B+C}{3}) \geq \dfrac{F(A)+F(B) +F(C)}{3}\)

log(sin(\(\pi/3\)) \(\geq \dfrac{log(sin(A)) +log(sin(B))+log(sin(C))}{3}\)

3 log(\(\dfrac{\sqrt{3}}{2}\)) \(\geq \log(\sin A+\sin B+\sin C)\)

Thus

\(\boxed{(\sin A. \sin B. \sin C) \leq \dfrac{3\sqrt3}{8}}\) Parth Lohomi · 2 years, 7 months ago

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@Parth Lohomi Rather directly apply on \(\sin x\). Sudeep Salgia · 2 years, 7 months ago

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(2)

\(A,B,C<\pi \Rightarrow\sin A, \sin B, \sin C >0\)

Now consider \(\sin A , \sin B , \sin C\).

We know that \(GM≤AM\), and hence the maximum value will occur only when \(GM=AM\), i.e. the terms are equal.

\(\therefore \sin A = \sin B = \sin C \Rightarrow A=B=C\)

As \(A+B+C=\pi\), we have \(A=B=C=\frac{\pi}{3}\)

Therefore, the maximum value of \(\sin A \sin B \sin C\) occurs at \(A=B=C=\frac{\pi}{3}\).

\(\therefore \sqrt[3]{\sin A \sin B \sin C} \le \frac{\sin A + \sin B + \sin C}{3}\)

\(\sqrt[3]{\sin A \sin B \sin C} \le \sin\frac{\pi}{3}\)

\(\sqrt[3]{\sin A \sin B \sin C} \le \frac{\sqrt{3}}{2}\)

\(\boxed{\sin A \sin B \sin C \le \frac{3\sqrt{3}}{8}}\) Omkar Kulkarni · 2 years, 7 months ago

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@Omkar Kulkarni Sorry I forgot to mention without AM-GM Parth Lohomi · 2 years, 7 months ago

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@Parth Lohomi Oh :P Omkar Kulkarni · 2 years, 7 months ago

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5) Let \( 1001 \leq k \leq 3001\)

A.M - H.M inequality -

\( \displaystyle \left(\sum_{k=1001}^{3001}k\right)\left( \sum_{k=1001}^{3001} \dfrac{1}{k}\right) > (2001)^2\)

\( \displaystyle \sum_{k=1001}^{3001} k = (2001)^2\)

Thus, \( \displaystyle \sum_{k=1001}^{3001} \dfrac{1}{k} > 1\)

\( M = \displaystyle \sum_{k=1001}^{3001} \dfrac{1}{k} < \dfrac{500}{1000} + \dfrac{500}{1500} + \dfrac{500}{2000} + \dfrac{1}{3001} \) ( grouped 500 terms simultaneously)

\( < \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{3000} = \dfrac{3851}{3000} < \dfrac{4}{3}\)

\( \boxed{1 < \dfrac{1}{1001} + .... + \dfrac{1}{3001} < \dfrac{4}{3}}\) U Z · 2 years, 7 months ago

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@U Z Good solution @megh choksi Parth Lohomi · 2 years, 7 months ago

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4) Good one , we can see RHS is inverse of LHS , thus

\( x^2 + 2ax + \dfrac{1}{16} = x\)

\( x^2 + x(2a - 1) + \dfrac{(2a - 1)^2}{4} = \dfrac{(2a - 1)^2}{4} + \dfrac{1}{16}\)

\( (x + (\dfrac{2a - 1}{2})^2 = \dfrac{(2a - 1)^2}{4} + \dfrac{1}{16}\)

\( \implies x = \dfrac{2a - 1}{2} \pm \sqrt{\dfrac{(2a - 1)^2}{4} + \dfrac{1}{16}}\)

Real roots depends on a.

From our equation - \(x^2 + x(2a - 1) - \dfrac{1}{16}\) , we can say for real roots discriminant should be greater than or equal to zero thus

\( (2a - 1)^2 - \dfrac{1}{16} \geq 0\)

\( \implies a \leq \dfrac{1}{4} ~or~ a \geq \dfrac{3}{4}\) U Z · 2 years, 7 months ago

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I think the last question should be -

\( 1 < \dfrac{1}{1001} + ...... + \dfrac{1}{3001} < \dfrac{4}{3}\) U Z · 2 years, 7 months ago

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@U Z Oops corrected! ! Parth Lohomi · 2 years, 7 months ago

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