Easy Peasy

Algebra Level 2

Let a , b , c , d > 0 a,b,c,d > 0 such that a + b + c + d = 1 a+b+c+d=1 . Find the minimum value of

1 4 a + 3 b + c + 1 4 d + 3 a + b + 1 4 c + 3 d + a + 1 4 b + 3 c + d \large \frac{1}{4a+3b+c}+\frac{1}{4d+3a+b}+\frac{1}{4c+3d+a}+\frac{1}{4b+3c+d}


The answer is 2.

Ojas Singh Malhi by Ojas Singh Malhi , 18, India

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2 solutions

Using Titu's lemma as suggested by @Md Zuhair

1 2 4 a + 3 b + c + 1 2 4 d + 3 a + b + 1 2 4 c + 3 d + a + 1 2 4 b + 3 c + d ≥ ( 1 + 1 + 1 + 1 ) 2 8 ( a + b + c + d ) = 2 \begin{aligned} \frac {1^2}{4a+3b+c} + \frac {1^2}{4d+3a+b} + \frac {1^2}{4c+3d+a} + \frac {1^2}{4b+3c+d} & \ge \frac {(1+1+1+1)^2}{8(a+b+c+d)} = \boxed{2} \end{aligned}

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Nicely done:)

Ojas Singh Malhi - 3 years, 8 months ago
Md Zuhair
Oct 2, 2017

simply applying Titu's lemma we get the answer

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