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.

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TopNewestFor 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.

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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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Yep, that's how I did it.

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first one is easy

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

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Never mind. Just realised that 4 was quite easy.

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

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

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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 breathingLog in to reply

For number 5, I'm trying to get the maximum value of 5/8

sqrt(2) + 1/3cube root(2)...Log in to reply