Maths Olympiad Prep

Library / /326 of 520

Algebra Difficulty 7.0 National olympiad Prove it

Example 1.14.12. Let a,b,c,da, b, c, d be positive real numbers with sum 4. Prove that
(1+a2)(1+b2)(1+c2)(1+d2)≥10493 \left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right)\left(1+d^{2}\right) \geq \frac{10^{4}}{9^{3}}

Solution

SOLUTION. We need to prove that
f(a)+f(b)+f(c)+f(d)≥4ln⁡10−3ln⁡9f(a)+f(b)+f(c)+f(d) \geq 4 \ln 10 - 3 \ln 9
where f(x)=ln⁡(1+x2) f(x) = \ln(1 + x^2) . Since
f′′(x)=2(1−x2)(1+x2)2f''(x) = \frac{2(1 - x^2)}{(1 + x^2)^2}
has exactly one positive real root x=1 x = 1 , we obtain by the SIP theorem that there exists a number p≤1 p \leq 1 for which
f(a)+f(b)+f(c)+f(d)≥3f(p)+f(4−3p)f(a) + f(b) + f(c) + f(d) \geq 3 f(p) + f(4 - 3p)

Denote
g(p)=3f(p)+f(4−3p)=3ln⁡(1+p2)+ln⁡(1+(4−3p)2)g(p) = 3 f(p) + f(4 - 3p) = 3 \ln(1 + p^2) + \ln(1 + (4 - 3p)^2)
then we get
g′(p)=6p1+p2−6(4−3p)1+(4−3p)2=24(p−1)2(3p−1)(1+p2)(1+(4−3p)2)g'(p) = \frac{6p}{1 + p^2} - \frac{6(4 - 3p)}{1 + (4 - 3p)^2} = \frac{24(p - 1)^2 (3p - 1)}{(1 + p^2)(1 + (4 - 3p)^2)}
and it is easy to conclude that
g(p)≥g(13)=4ln⁡10−3ln⁡9g(p) \geq g\left(\frac{1}{3}\right) = 4 \ln 10 - 3 \ln 9
as desired. The equality holds for a=b=c=13,d=3 a = b = c = \frac{1}{3}, d = 3 or permutations.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.