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
4 years ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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 - 4 years ago

Log in to reply

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.

Log in to reply

Yep, that's how I did it.

Log in to reply

first one is easy

Mardokay Mosazghi - 4 years ago

Log in to reply

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

Ayush Garg - 2 years, 12 months ago

Log in to reply

Never mind. Just realised that 4 was quite easy.

Ayush Garg - 2 years, 11 months ago

Log in to reply

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

Dinesh Chavan - 3 years, 11 months ago

Log in to reply

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

Samuraiwarm Tsunayoshi - 3 years, 11 months ago

Log in to reply

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

Log in to reply

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

John Ashley Capellan - 4 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...