RMO inequalities DPP 1

October 02, 2025

First focus on a^5 + b^5.
WLOG a >= b
=> a^2 >= b^2
=> a^3 >= b^3
By rearrangement inequality:
a^2.a^3 + b^2.b^3 >= a^2.b^3 + b^2.a^3
=> a^5 + b^5 >= a^2.b^3 + b^2.a^3

Add ab to both sides to get the first denominator:
a^5+ b^5 + ab >= a^2.b^3 + b^2.a^3 + ab = ab(1 + a^2.b + a.b^2)
Cross multiplication:
1/(1 + a^2.b + a.b^2) >= ab/(a^5+ b^5 + ab)
Now we can see that RHS is the first term of the original question expression.
Now if we put abc = 1 in LHS:
then LHS = abc/(abc + a^2.b + a.b^2)
Divide numerator/denominator by ab:
LHS = c/(c + a + b)
So:
c/(c + a + b) >= ab/(a^5+ b^5 + ab)
Similarly:
b/(c + a + b) >= ac/(a^5+ c^5 + ac)
a/(c + a + b) >= bc/(b^5+ c^5 + bc)

Add all:
LHS = 1
RHS is the original question expression.
=>
1 >= original expression.
Hence proved.

In the above question we used Cauchy Shwarz Inequality.

We will use Titu's Lemma here.

Above to prove that x^4 + 2x >= 3x^2
We can use A.M. G.M. as well.
Apply AM GM on
x^4, x, x to get this.

First part of this solution is not correct. But the point is that we can't get a meaningful bound on S without dividing 1+y, 1+z by 8.

Take cases:
1 variable 0, 2 adjacent 0(a,b), 2 non-adjacent 0(a,c), >=3 variables zero.
You will be easily able to prove that inequality holds in these cases.

Equality is achieved at a = b = c = d = e = 1.
From there it's slightly difficult to prove.

Another way: