High school inequality proof Given ABC = 1, and a, B, C are real numbers, it is proved that 1 / A + 1 / B + 1 / C + 3 / (a + B + C) > = 4

High school inequality proof Given ABC = 1, and a, B, C are real numbers, it is proved that 1 / A + 1 / B + 1 / C + 3 / (a + B + C) > = 4

This problem is an Olympic problem in middle mathematics (senior high school): A, B, C are all positive numbers only! (counter example can be given) the original solution is made with the adjustment method, here we seriously recommend the algebraic identity deformation + basic inequality method!

Prove an inequality
(x+1/x)^n+2>=x^n+1/(x^n)+2^n
n=1,2,3,4.

Solving binomial theorem
(x+1/x)^n=x^n+x^(n-2)+…… +x^2+1+x^(-2)+…… +X ^ (- N + 2) + X (- n) (binomial theorem)
So (x + 1 / x) ^ n - (x ^ n + 1 / x ^ n)
=x^(n-2)+…… +x^2+1+x^(-2)+…… +x^(-n+2)
=(x+1/x)^(n-2)
According to the inequality x + 1 / X ≥ 2
So (x + 1 / x) ^ (n-2) ≥ 2 ^ (n-2)
(x+1/x)^n-(x^n+1/x^n)≥2^n-2
That is, (x + 1 / x) ^ n + 2 ≥ (x ^ n + 1 / x ^ n) + 2 ^ n

A proof of inequality in Senior High School
If a > 0, b > 0, a + B = 1, it is proved that (a + 1 / a) × (B + 1 / b) ≥ 25 / 4

The original formula is equal to (AB + 1 / AB) + (A / B + B / a), and it is divided into two groups to find the minimum value. For the first group, obviously 0