Track / Stage 5 / 337 of 400 #937 of 2000
Problem 937 AIME late Difficulty 5.8 Prove it
[ [Quadratic inequalities (several variables) . ] [ Ordering by increasing (decreasing) ] \left[\begin{array}{l}\text { [Quadratic inequalities (several variables) }.] \\ {[\text { Ordering by increasing (decreasing) }}\end{array}\right] [ [Quadratic inequalities (several variables) . ] [ Ordering by increasing (decreasing) ]
Prove the inequality ( a + b + c + d + 1 ) 2 ≥ 4 ( a 2 + b 2 + c 2 + d 2 ) (a+b+c+d+1)^{2} \geq 4\left(a^{2}+b^{2}+c^{2}+d^{2}\right) ( a + b + c + d + 1 ) 2 ≥ 4 ( a 2 + b 2 + c 2 + d 2 ) a , b , c , d ∈ [ 0 , 1 ] a, b, c, d \in[0,1] a , b , c , d ∈ [ 0 , 1 ]
This one wants a proof. Work it on paper, read the official solution, then mark
yourself honestly — the ladder only means something if the record is true.
I solved it I didn't Skip
Official solution We can assume that a ≥ b ≥ c ≥ d a \geq b \geq c \geq d a ≥ b ≥ c ≥ d
( a + b + c + d + 1 ) 2 = ( a 2 + b 2 + c 2 + d 2 ) + 2 ( a b + a c + a d + b c + b d + c d ) + 2 ( a + b + c + d ) + 1 ≥ (a+b+c+d+1)^{2}=\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+2(a b+a c+a d+b c+b d+c d)+2(a+b+c+d)+1 \geq ( a + b + c + d + 1 ) 2 = ( a 2 + b 2 + c 2 + d 2 ) + 2 ( ab + a c + a d + b c + b d + c d ) + 2 ( a + b + c + d ) + 1 ≥
≥ ( a 2 + b 2 + c 2 + d 2 ) + 2 ( b 2 + c 2 + d 2 + c 2 + d 2 + d 2 ) + 2 ( a 2 + b 2 + c 2 + d 2 ) + a 2 ≥ 4 ( a 2 + b 2 + c 2 + d 2 ) \geq\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+2\left(b^{2}+c^{2}+d^{2}+c^{2}+d^{2}+d^{2}\right)+2\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+a^{2} \geq 4\left(a^{2}+b^{2}+c^{2}+d^{2}\right) ≥ ( a 2 + b 2 + c 2 + d 2 ) + 2 ( b 2 + c 2 + d 2 + c 2 + d 2 + d 2 ) + 2 ( a 2 + b 2 + c 2 + d 2 ) + a 2 ≥ 4 ( a 2 + b 2 + c 2 + d 2 )
Submit a comment
← Previous All problems Next →
Source: NuminaMath-1.5 ,
licensed Apache-2.0 .
Statement and solution reproduced as published; topic, difficulty and ordering added
by this site.