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

Revision as of 23:31, 26 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,

$\frac{a. a +b. b +c. c +d. d}{a+b+c+d} \ge (a^a b^b c^c d^d)^{\frac{1}{a+b+c+d}}$

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

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 $a=b=c=d=\frac{1}{4}$

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

@@ Line 15: / Line 15: @@
 Now notice that ,
+    <cmath>a+2b+3c+4d \text{ will be less then the following expression (and reason is written on right)} </cmath>
+    <cmath>a+2b+3c+3d ,\text{as} d\le b</cmath>
+    <cmath>3a+3b+3c+d,              \text{as}  d\le a</cmath>
+    <cmath>3a+b+3c+3d , \text{as}  b+d\le 2a </cmath>
+  <cmath>3a +3b +c +3d , \text{as} 2c+d \le 2a+b </cmath>
-\[
-    a+2b+3c+4d \le
-\begin{cases}
-    a+3b+3c+3d,& \text{as }  d\le b\\
-a+3b+3c+d,              &\text{as}  d\le a \\
-a+b+3c+3d ,& \text{as}
- b+d\le 2a \\
-a +3b +c +3d ,& \text{as}
-c+d \le 2a+b
-\end{cases}
-\]
 So, We get ,