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My discussions $$1$$

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, 5 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

$$\boxed{(\sin A. \sin B. \sin C) \leq \dfrac{3\sqrt3}{8}}$$ · 2 years, 5 months ago

Rather directly apply on $$\sin x$$. · 2 years, 5 months ago

(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 $$A=B=C=\frac{\pi}{3}$$.

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

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

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

$$\boxed{\sin A \sin B \sin C \le \frac{3\sqrt{3}}{8}}$$ · 2 years, 5 months ago

Sorry I forgot to mention without AM-GM · 2 years, 5 months ago

Oh :P · 2 years, 5 months ago

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)

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

$$\boxed{1 < \dfrac{1}{1001} + .... + \dfrac{1}{3001} < \dfrac{4}{3}}$$ · 2 years, 5 months ago

Good solution @megh choksi · 2 years, 5 months ago

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

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

$$\implies a \leq \dfrac{1}{4} ~or~ a \geq \dfrac{3}{4}$$ · 2 years, 5 months ago

I think the last question should be -

$$1 < \dfrac{1}{1001} + ...... + \dfrac{1}{3001} < \dfrac{4}{3}$$ · 2 years, 5 months ago