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(11)

If a1,a2,a3...............a24a_1,a_2,a_3...............a_{24} are such that

a2a1=a3a2=............=a24a23\dfrac{a_2}{a_1}= \dfrac{a_3}{a_2}=............ =\dfrac{a_{24}}{a_{23}}

and i=124\displaystyle\sum_{i=1}^{24}ai=2a_i = 2

i=124\displaystyle\sum_{i=1}^{24}1ai=1\dfrac{1}{a_i} = 1

Find i=124\displaystyle\prod_{i=1}^{24}aia_i

(22)

For an acute-angled ABC\triangle{ABC} ] 0,π/20,\pi/2] , proove that (without using AM-GM)

sin(A).sin(B).sin(C)338\sin(A). \sin(B). \sin(C) \leq \dfrac{3\sqrt3}{8}

(33)

Find the largest constant kk such that

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

(44)

Find the real roots of the equation

x2+2ax+116=a+a2+x116x^2+2ax+\dfrac{1}{16} = - a+\sqrt{a^2+x-\dfrac{1}{16}}

(55)

Proove that

1<11001+11002+............+13001<431 <\dfrac{1}{1001} + \dfrac{1}{1002} +............ +\dfrac{1}{3001} < \dfrac{4}{3}


Hope you will enjoy them!!

Note by Parth Lohomi
4 years, 8 months ago

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Solution for( 2)

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

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

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

Apply jensens inequality

F(A+B+C3)F(A)+F(B)+F(C)3F(\dfrac{A+B+C}{3}) \geq \dfrac{F(A)+F(B) +F(C)}{3}

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

3 log(32\dfrac{\sqrt{3}}{2}) log(sinA+sinB+sinC)\geq \log(\sin A+\sin B+\sin C)

Thus

(sinA.sinB.sinC)338\boxed{(\sin A. \sin B. \sin C) \leq \dfrac{3\sqrt3}{8}}

Parth Lohomi - 4 years, 8 months ago

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Rather directly apply on sinx\sin x.

Sudeep Salgia - 4 years, 8 months ago

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(2)

A,B,C<πsinA,sinB,sinC>0A,B,C<\pi \Rightarrow\sin A, \sin B, \sin C >0

Now consider sinA,sinB,sinC\sin A , \sin B , \sin C.

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

sinA=sinB=sinCA=B=C\therefore \sin A = \sin B = \sin C \Rightarrow A=B=C

As A+B+C=πA+B+C=\pi, we have A=B=C=π3A=B=C=\frac{\pi}{3}

Therefore, the maximum value of sinAsinBsinC\sin A \sin B \sin C occurs at A=B=C=π3A=B=C=\frac{\pi}{3}.

sinAsinBsinC3sinA+sinB+sinC3\therefore \sqrt[3]{\sin A \sin B \sin C} \le \frac{\sin A + \sin B + \sin C}{3}

sinAsinBsinC3sinπ3\sqrt[3]{\sin A \sin B \sin C} \le \sin\frac{\pi}{3}

sinAsinBsinC332\sqrt[3]{\sin A \sin B \sin C} \le \frac{\sqrt{3}}{2}

sinAsinBsinC338\boxed{\sin A \sin B \sin C \le \frac{3\sqrt{3}}{8}}

Omkar Kulkarni - 4 years, 8 months ago

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Sorry I forgot to mention without AM-GM

Parth Lohomi - 4 years, 8 months ago

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Oh :P

Omkar Kulkarni - 4 years, 8 months ago

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5) Let 1001k3001 1001 \leq k \leq 3001

A.M - H.M inequality -

(k=10013001k)(k=100130011k)>(2001)2 \displaystyle \left(\sum_{k=1001}^{3001}k\right)\left( \sum_{k=1001}^{3001} \dfrac{1}{k}\right) > (2001)^2

k=10013001k=(2001)2 \displaystyle \sum_{k=1001}^{3001} k = (2001)^2

Thus, k=100130011k>1 \displaystyle \sum_{k=1001}^{3001} \dfrac{1}{k} > 1

M=k=100130011k<5001000+5001500+5002000+13001 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)

<12+13+14+15+13000=38513000<43 < \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{3000} = \dfrac{3851}{3000} < \dfrac{4}{3}

1<11001+....+13001<43 \boxed{1 < \dfrac{1}{1001} + .... + \dfrac{1}{3001} < \dfrac{4}{3}}

U Z - 4 years, 8 months ago

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Good solution @megh choksi

Parth Lohomi - 4 years, 8 months ago

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I think the last question should be -

1<11001+......+13001<43 1 < \dfrac{1}{1001} + ...... + \dfrac{1}{3001} < \dfrac{4}{3}

U Z - 4 years, 8 months ago

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Oops corrected! !

Parth Lohomi - 4 years, 8 months ago

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

x2+2ax+116=x x^2 + 2ax + \dfrac{1}{16} = x

x2+x(2a1)+(2a1)24=(2a1)24+116 x^2 + x(2a - 1) + \dfrac{(2a - 1)^2}{4} = \dfrac{(2a - 1)^2}{4} + \dfrac{1}{16}

(x+(2a12)2=(2a1)24+116 (x + (\dfrac{2a - 1}{2})^2 = \dfrac{(2a - 1)^2}{4} + \dfrac{1}{16}

    x=2a12±(2a1)24+116 \implies x = \dfrac{2a - 1}{2} \pm \sqrt{\dfrac{(2a - 1)^2}{4} + \dfrac{1}{16}}

Real roots depends on a.

From our equation - x2+x(2a1)116x^2 + x(2a - 1) - \dfrac{1}{16} , we can say for real roots discriminant should be greater than or equal to zero thus

(2a1)21160 (2a - 1)^2 - \dfrac{1}{16} \geq 0

    a14 or a34 \implies a \leq \dfrac{1}{4} ~or~ a \geq \dfrac{3}{4}

U Z - 4 years, 8 months ago

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