Inequality in four variables

To prove that

(5a12+b3+c6+d12)2≤5a212+b23+c26+d212 \left( \frac{5a}{12}+ \frac{b}{3}+ \frac{c}{6}+ \frac{d}{12} \right) ^2 \le \frac{5a^2}{12}+ \frac{b^2}{3}+ \frac{c^2}{6}+ \frac{d^2}{12}

a,b,c,d∈R+ a,b,c,d \in \mathbb{R^+}

Shared by my friend Narmad Raval

#Algebra #Inequality

Note by Megh Parikh
7 years, 1 month ago

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Comments

Expand To Get

536(a−b)2+572(a−c)2+5144(a−d)2+118(b−c)2+136(b−d)2+172(c−d)2≥0\frac {5}{36} (a-b)^2 + \frac {5}{72} (a-c)^2 + \frac {5}{144} (a-d)^2 + \frac {1}{18} (b-c)^2 + \frac {1}{36} (b-d)^2 + \frac {1}{72} (c-d)^2 \geq 0

ANANT Chhajwani - 7 years, 1 month ago

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Thanks .

I dont know how I could miss that.

Megh Parikh - 7 years, 1 month ago

By Cauchy-Schwarz inequality,

(512+13+16+112)(5a212+b23+c26+d212)≥(5a12+b3+c6+d12)2 \left( \dfrac{5}{12} + \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{12} \right)\left(\dfrac{5a^2}{12} + \dfrac{b^2}{3} + \dfrac{c^2}{6} + \dfrac{d^2}{12}\right) \geq \left( \dfrac{5a}{12} + \dfrac{b}{3} + \dfrac{c}{6} + \dfrac{d}{12} \right)^2

⇒(5a212+b23+c26+d212)≥(5a12+b3+c6+d12)2 \Rightarrow \left(\dfrac{5a^2}{12} + \dfrac{b^2}{3} + \dfrac{c^2}{6} + \dfrac{d^2}{12}\right) \geq \left( \dfrac{5a}{12} + \dfrac{b}{3} + \dfrac{c}{6} + \dfrac{d}{12} \right)^2

Siddhartha Srivastava - 7 years, 1 month ago
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