An old IMO shortlist

Algebra Level 5

It is given that a , b , c , d > 0 a, b, c, d > 0 and a b + b c + c d + d a = 1 ab + bc + cd + da = 1 . Find the minimum value of

c y c a 3 b + c + d \sum_{cyc}\frac{a^{3}}{b + c + d}


The answer is 0.3333333333333.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Raushan Sharma
Apr 19, 2016

c y c a 3 b + c + d \sum_{cyc}\frac{a^{3}}{b + c + d}

= c y c a 4 a b + a c + a d =\sum_{cyc}\frac{a^{4}}{ab + ac + ad}

( a 2 + b 2 + c 2 + d 2 ) 2 2 a b + 2 a c + 2 a d + 2 b c + 2 b d + 2 c d \geq \frac{(a^2+b^2+c^2+d^2)^2}{2ab+2ac+2ad+2bc+2bd+2cd} [By Tittu's Lemma]

( a 2 + b 2 + c 2 + d 2 ) 2 a 2 + b 2 + a 2 + c 2 + a 2 + d 2 + b 2 + c 2 + b 2 + d 2 + c 2 + d 2 \geq \frac{(a^2+b^2+c^2+d^2)^2}{a^2+b^2+a^2+c^2+a^2+d^2+b^2+c^2+b^2+d^2+c^2+d^2}

= ( a 2 + b 2 + c 2 + d 2 ) 2 3 ( a 2 + b 2 + c 2 + d 2 ) =\frac{(a^2+b^2+c^2+d^2)^2}{3(a^2+b^2+c^2+d^2)}

= a 2 + b 2 + c 2 + d 2 3 =\frac{a^2+b^2+c^2+d^2}{3}

1 3 \geq \frac{1}{3}

What I used in the last step is : ( a 2 + b 2 ) + ( b 2 + c 2 ) + ( c 2 + d 2 ) + ( d 2 + a 2 ) 2 ( a b + b c + c d + d a ) = 2 (a^2+b^2)+(b^2+c^2)+(c^2+d^2)+(d^2+a^2) \geq 2(ab+bc+cd+da) = 2

So, a 2 + b 2 + c 2 + d 2 1 a^2+b^2+c^2+d^2 \geq 1

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...