Only use any the following.

- Basic inequalities
- AM-GM-HM
- Holder
- Cauchy Schwarz (Titu's lemma is acceptable)
- Weighted AM-GM

1.) Let \(a,b,c\) be positive real numbers such that \(abc = 1\). Prove that

\[\frac{4}{(a+2)^{2}+4b^{2}} + \frac{4}{(b+2)^{2}+4c^{2}} + \frac{4}{(c+2)^{2}+4a^{2}} \leq 1\]

2.) Find all positive real solutions \(a,b,c\) such that

\[ \begin{cases} 10a^{3}+b^{3} = 7ab + 20bc - 11 \\ b^{3}+20c^{3} = 7bc + 30ca - 20 \\ 34c^{3}+44a^{3} = 51ca + 20ab - 50 \\ \end{cases}\]

3.) Let \(a,b,c\) be positive real numbers such that \(abc = 1\). Prove that

\[a^{3}b+b^{3}c+c^{3}a \geq a^{2/5}b^{3/5} + b^{2/5}c^{3/5} + c^{2/5}a^{3/5}\]

4.) Let \(a,b,c\) be positive real numbers such that \(a+b+c+abc = 4\). Prove that

\[\left(\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\right)^{2}(ab+bc+ca) \geq \frac{1}{2}(4-abc)^{3}\]

5.) Let \(a,b,c\) be positive real numbers. Prove that

\[\frac{a^{9}}{bc}+\frac{b^{9}}{ca}+\frac{c^{9}}{ab} + \frac{3}{abc} \geq a^{4}+b^{4}+c^{4}+3\]

Check out all my notes and stuffs for more problems!

## Comments

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TopNewestI like the kind of questions where a suitably application of AM-GM gives the resultant inequality, since that relies on your ability to spot / identify patterns.

For example, here is how to do 3:

Apply AM-GM to \( 2 a^3 b + 2 b^3 c + c^3 a \). – Calvin Lin Staff · 2 years, 9 months ago

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Muirheaders gonna Muirhead Muirhead Muirhead – Daniel Liu · 2 years, 8 months ago

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– Calvin Lin Staff · 2 years, 8 months ago

Right. Muirhead is just a fancy term for bunching AM-GM.Log in to reply

– Daniel Liu · 2 years, 8 months ago

In addition, graders hate it because it involves no ingenuity at all.Log in to reply

Sorry I miss typed no.4 here. It is \(a+b+c+abc = 4\) instead of \(1\). – Samuraiwarm Tsunayoshi · 2 years, 8 months ago

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Is it necessary to use just the ineqs u stated? – Dinesh Chavan · 2 years, 9 months ago

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– Samuraiwarm Tsunayoshi · 2 years, 9 months ago

Yup.Log in to reply