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

2.) Let \(n\) be positive integers \(n \geq 2\) and \(x_{1},y_{1},x_{2},y_{2},...,x_{n},y_{n}\) be positive real numbers such that

\[\sum\limits_{i = 1}^{n} x_{i} \geq \sum\limits_{i = 1}^{n} x_{i}y_{i}\]

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

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

\[a^{3}+b^{3}+c^{3} \geq \frac{1}{4}(a+b+c)^{3} - 6abc\]

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

\[abcd(a+b+c+d)(ab^{1/2}c^{1/3}d^{1/4}) \leq \frac{1}{\sqrt{6}}(a^{4}b+b^{4}c+c^{4}d+d^{4}a)\]

5.) Let \(a\geq 2, b\geq 6, c \geq 12\), find and show the maximum value of

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

and find the conditions of \(a,b,c\) to be maximum.

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

Note by Samuraiwarm Tsunayoshi
3 years, 6 months 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 - 3 years, 6 months ago

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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.

Samuraiwarm Tsunayoshi - 3 years, 6 months ago

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Yep, that's how I did it.

Samuraiwarm Tsunayoshi - 3 years, 6 months ago

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first one is easy

Mardokay Mosazghi - 3 years, 6 months ago

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

Ayush Garg - 2 years, 6 months ago

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Never mind. Just realised that 4 was quite easy.

Ayush Garg - 2 years, 6 months ago

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

Dinesh Chavan - 3 years, 6 months ago

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

Samuraiwarm Tsunayoshi - 3 years, 6 months ago

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Trying #5...

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

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

Let \(x = a-2, y = b-6, z = c-12\) such that \(x,y,z \geq 0\).

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

By AM-GM; \[\displaystyle \frac{\sqrt{x}}{x+2} \leq \frac{\sqrt{x}}{2\sqrt{2x}} = \frac{1}{2\sqrt{2}}\]

Similarly, \[\displaystyle \frac{\sqrt[3]{y}}{y+3+3} \leq \frac{\sqrt[3]{y}}{3\sqrt[3]{y\times3^{2}}} = \frac{1}{3\times 3^{2/3}}\]

\[\displaystyle \frac{\sqrt[4]{z}}{z+4+4+4} \leq \frac{\sqrt[4]{z}}{4\sqrt[4]{z\times4^{3}}} = \frac{1}{4\times 4^{3/4}}\]

Sum all these up and we get

\[\displaystyle \frac{bc\sqrt{a-2} + ca\sqrt[3]{b-6} + ab\sqrt[4]{c-12}}{abc} \leq \frac{1}{2\sqrt{2}}+\frac{1}{3\times 3^{2/3}}+\frac{1}{4\times 4^{3/4}}\] ~~~

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

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

Samuraiwarm Tsunayoshi - 3 years, 6 months 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 - 3 years, 6 months ago

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