Difference between revisions of "2020 IMO Problems/Problem 2"

Revision as of 19:52, 27 September 2020

Problem 2. The real numbers $a, b, c, d$ are such that $a\ge b \ge c\ge d > 0$ and $a+b+c+d=1$ . Prove that $(a+2b+3c+4d)a^a b^bc^cd^d<1$

Using Weighted AM-GM we get

$\frac{a\cdot a +b\cdot b +c\cdot c +d\cdot d}{a+b+c+d} \ge \sqrt[a+b+c+d]{a^a b^b c^c d^d}$

$\implies a^a b^b c^c d^d \le a^2 +b^2 +c^2 +d^2$

So, $(a+2b+3c+4d) a^ab^bc^c \le (a+2b+3c+4d)(a^2+b^2+c^2+d^2)$

Now notice that

So, we get $(a+2b+3c+4d)(a^2+b^2+c^2+d^2)$ $= a^2(a+2b+3c+4d)+b^2(a+2b+3c+4d)+c^2 (a+2b+3c+4d) +d^2 (a+2b+3c+4d)$

$\le a^2(a+3b+3c+3d)+b^2(3a+b+3c+3d)+c^2 (3a+3b+c+3d) +d^2 (3a+3b+3c+d)$

$=(a+b+c+d)^3 =1$

Now, for equality we must have $a=b=c=d=\frac{1}{4}$

In that case we get $(a+2b+3c+4d) a^ab^bc^c \le (a+2b+3c+4d)(a^2+b^2+c^2+d^2) =\frac{5}{8} <1$

~ftheftics

@@ Line 8: / Line 8: @@
 Using Weighted AM-GM we get
-<cmath>\frac{a. a +b. b +c. c +d. d}{a+b+c+d} \ge (a^a b^b c^c d^d)^{\frac{1}{a+b+c+d}}</cmath>
+<cmath>\frac{a\cdot a +b\cdot b +c\cdot c +d\cdot d}{a+b+c+d} \ge \sqrt[a+b+c+d]{a^a b^b c^c d^d}</cmath>
 <cmath>\implies a^a b^b c^c d^d \le a^2 +b^2 +c^2 +d^2</cmath>