Inequality Problem Set IV

Inequality Problem Set IV
Maths Olympiad Preparation

22nd February 2024




Problems & Solutions

Let a, b & c ∈ R+ with a + b + c = 1. Prove that
X
3 (b + ca)2 (c + ab)2 ≥ 4(a + bc)(b + ca)(c + ab).
cyc



q
Claim: cyc (b+ca)(c+ab)
P
a+bc = 2.
Proof: We have
a + bc = 1 − b − c + bc = (1 − b)(1 − c).

Therefore,
r
X (b + ca)(c + ab)
= 3 − (a + b + c) = 2.
cyc
a + bc

This proves our Claim.
Now, Power Mean Inequality gives us

(x + y + z)2 X (b + ca)(c + ab) 4
x2 + y 2 + z 2 ≥ =⇒ ≥ ,
3 cyc
a + bc 3

X
i.e. 3 (b + ca)2 (c + ab)2 ≥ 4(a + bc)(b + ca)(c + ab).
cyc


Equality holds iff a = b = c = 31 .
QED.




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In ∆ABC, prove that

A B C s3
cos4 + cos4 + cos4 ≤ .
2 2 2 2abc


We make the Ravi substitution as follows

a = y + z, b = z + x & c = x + y.

Then, we have
a+b+c
s= = x + y + z.
2
And
A s(s − a) x(x + y + z)
cos2 = = .
2 bc (y + z)(z + x)
Therefore, we need to prove that

A B C s3
cos4 + cos4 + cos4 ≤
2 2 2 2abc
X  x2 (x + y + z)2 2 (x + y + z)3
i.e. ≤
cyc
(z + x)(x + y) 2(y + z)(z + x)(x + y)
X
i.e. 2 x2 (y + z)2 ≤ (x + y + z)(y + z)(z + x)(x + y)
cyc
X X
i.e. 4 y 2 z 2 + 4xyz(x + y + z) ≤ 4xyz(x + y + z) + 2 y2 z2
cyc cyc
X
+ yz(y 2 + z 2 )
cyc

i.e. 2(y 2 z 2 + z 2 x2 + x2 y 2 ) ≤ yz(y 2 + z 2 ) + zx(z 2 + x2 ) + xy(x2 + y 2 ).

Which is clearly true by AM-GM.
Equality holds iff a = b = c.
QED.




2 Maths Olympiad Preparation