Sequence and series (10)

Algebra Level 3

For positive reals a a , b b , and c c , what is the minimum value of a 3 4 b + b 8 c 2 + 1 + c 2 a \dfrac{a^3}{4b} + \dfrac{b}{8c^2} + \dfrac{1+c}{2a} ?

5 4 \dfrac{5}{4} 11 4 \dfrac{11}{4} 30 4 \dfrac{30}{4} 21 4 \dfrac{21}{4}

Adhiraj Dutta by Adhiraj Dutta , 18, India

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1 solution

Vilakshan Gupta
Apr 25, 2020

a 3 4 b + b 8 c 2 + 1 + c 2 a = a 3 4 b + b 8 c 2 + 1 2 a + c 4 a + c 4 a ≥ 5 ( a 3 4 b × b 8 c 2 × 1 2 a × c 4 a × c 4 a ) 1 / 5 = 5 4 \dfrac{a^3}{4b}+\dfrac{b}{8c^2}+\dfrac{1+c}{2a}=\dfrac{a^3}{4b}+\dfrac{b}{8c^2}+\dfrac{1}{2a}+\dfrac{c}{4a}+\dfrac{c}{4a} \geq 5 \left(\dfrac{a^3}{4b} \times \dfrac{b}{8c^2} \times \dfrac{1}{2a} \times \dfrac{c}{4a} \times \dfrac{c}{4a} \right)^{1/5} = \boxed{\dfrac{5}{4}} (By using AM-GM Inequality ).

Equality occurs when a = c = 2 , b = 8 a=c=2 , b=8 .

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