Inequalities (1st math Thailand POSN 2014)

Only use these following inequalities:

  • A.M.-G.M.-H.M.
  • Cauchy Schwarz
  • Weighted A.M-G.M. with weight sum = 1

1.) Let a,b,ca,b,c be positive real numbers such that (a+1a)(b+1b)(c+1c)abc=8\displaystyle \frac{\left(\displaystyle a+ \frac{1}{a}\right)\left(\displaystyle b+ \frac{1}{b}\right)\left(\displaystyle c+ \frac{1}{c}\right)}{abc} = 8 Prove that abc1abc \geq 1.

2.) Let nn be positive integers n2n \geq 2 and x1,y1,x2,y2,...,xn,ynx_{1},y_{1},x_{2},y_{2},...,x_{n},y_{n} be positive real numbers such that

i=1nxii=1nxiyi\sum\limits_{i = 1}^{n} x_{i} \geq \sum\limits_{i = 1}^{n} x_{i}y_{i}

Prove that i=1nxii=1nxiyi\sum\limits_{i = 1}^{n} x_{i} \leq \sum\limits_{i = 1}^{n} \displaystyle \frac{x_{i}}{y_{i}}

3.) Let a,b,ca,b,c be positive real numbers, prove that

a3+b3+c314(a+b+c)36abca^{3}+b^{3}+c^{3} \geq \frac{1}{4}(a+b+c)^{3} - 6abc

4.) Let a,b,c,da,b,c,d be positive real numbers and 36a+4b+4c+3d=2536a+4b+4c+3d = 25, prove that

abcd(a+b+c+d)(ab1/2c1/3d1/4)16(a4b+b4c+c4d+d4a)abcd(a+b+c+d)(ab^{1/2}c^{1/3}d^{1/4}) \leq \frac{1}{\sqrt{6}}(a^{4}b+b^{4}c+c^{4}d+d^{4}a)

5.) Let a2,b6,c12a\geq 2, b\geq 6, c \geq 12, find and show the maximum value of

bca2+cab63+abc124abc\displaystyle \frac{bc\sqrt{a-2} + ca\sqrt[3]{b-6} + ab\sqrt[4]{c-12}}{abc}

and find the conditions of a,b,ca,b,c to be maximum.

This is the part of Thailand 1st round math POSN problems.

Note by Samuraiwarm Tsunayoshi
5 years ago

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For number 1, using the fact that a + (1/a) >= 2, this also applies for b and c. Hence, [a + (1/a)][b + (1/b)][c + (1/c)] >= 8. Hence, 8abc >= 8 and abc >= 1.

John Ashley Capellan - 5 years ago

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Yep, that's how I did it.

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This is the easiest question because professor let everyone get at least 1 problem to bring some joy of solving the hardest topic of all.

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first one is easy

Mardokay Mosazghi - 5 years ago

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Can someone tell me how to solve 4? I solved the rest.

Ayush Garg - 4 years ago

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Never mind. Just realised that 4 was quite easy.

Ayush Garg - 4 years ago

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For number 5, I'm trying to get the maximum value of 5/8sqrt(2) + 1/3 cube root(2)...

John Ashley Capellan - 5 years ago

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Trying #5...

bca2+cab63+abc124abc\displaystyle \frac{bc\sqrt{a-2} + ca\sqrt[3]{b-6} + ab\sqrt[4]{c-12}}{abc}

=a2a+b63b+c124c= \displaystyle \frac{\sqrt{a-2}}{a} + \frac{\sqrt[3]{b-6}}{b} + \frac{\sqrt[4]{c-12}}{c}

Let x=a2,y=b6,z=c12x = a-2, y = b-6, z = c-12 such that x,y,z0x,y,z \geq 0.

The expression becomes =xx+2+y3y+6+z4z+12= \displaystyle \frac{\sqrt{x}}{x+2} + \frac{\sqrt[3]{y}}{y+6} + \frac{\sqrt[4]{z}}{z+12}

By AM-GM; xx+2x22x=122\displaystyle \frac{\sqrt{x}}{x+2} \leq \frac{\sqrt{x}}{2\sqrt{2x}} = \frac{1}{2\sqrt{2}}

Similarly, y3y+3+3y33y×323=13×32/3\displaystyle \frac{\sqrt[3]{y}}{y+3+3} \leq \frac{\sqrt[3]{y}}{3\sqrt[3]{y\times3^{2}}} = \frac{1}{3\times 3^{2/3}}

z4z+4+4+4z44z×434=14×43/4\displaystyle \frac{\sqrt[4]{z}}{z+4+4+4} \leq \frac{\sqrt[4]{z}}{4\sqrt[4]{z\times4^{3}}} = \frac{1}{4\times 4^{3/4}}

Sum all these up and we get

bca2+cab63+abc124abc122+13×32/3+14×43/4\displaystyle \frac{bc\sqrt{a-2} + ca\sqrt[3]{b-6} + ab\sqrt[4]{c-12}}{abc} \leq \frac{1}{2\sqrt{2}}+\frac{1}{3\times 3^{2/3}}+\frac{1}{4\times 4^{3/4}} ~~~

Equality occurs if and only if x=2,y=3,z=4x=2,y=3,z=4 which means a=4,b=9,c=16a=4,b=9,c=16 ~~~

I don't have much time for doing these problems right now. I have to study math from POSN for whole October and I'm freaking busy reviewing lots of stuffs. heavy breathing

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3rd one is preety easy. Do u still need it?

Dinesh Chavan - 4 years, 12 months ago

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Yup I'm still stuck till now. =_="

Samuraiwarm Tsunayoshi - 4 years, 12 months ago

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