The minimum value of f(x)=(x−2)10+(x−2)11+3(x−2)12+(x−2)15+(x−2)23+(x−2)25(x−2)15, ∀ x>2

f(x)=(x−2)−5+(x−2)−4+3(x−2)−4+3(x−2)−3+1+(x−2)8+(x−2)10 Say x-2 = t (> 0)f(x)=t−5+t−4+3t−3+1+t8+t10.Now apply AM≥GMt−5+t−4+t−3+t−3+t−3+1+t8+t108≥(t−5.t−4.t−3.t−3.1.t8.t10)1/8⇒f(x)≥8∀x>2min value of f(x) is 8

The minimum value of f(x)=(x−2)10+(x−2)11+3(x−2)12+(x−2)15+(x−2)23+(x−2)25(x−2)15, ∀ x>2

f(x)=(x−2)−5+(x−2)−4+3(x−2)−4+3(x−2)−3+1+(x−2)8+(x−2)10 Say x-2 = t (> 0)f(x)=t−5+t−4+3t−3+1+t8+t10.Now apply AM≥GMt−5+t−4+t−3+t−3+t−3+1+t8+t108≥(t−5.t−4.t−3.t−3.1.t8.t10)1/8⇒f(x)≥8∀x>2min value of f(x) is 8

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The minimum value of $f (x) = \frac{{(x - 2)}^{10} + {(x - 2)}^{11} + 3 {(x - 2)}^{12} + {(x - 2)}^{15} + {(x - 2)}^{23} + {(x - 2)}^{25}}{{(x - 2)}^{15}}, \forall x > 2$

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Detailed Solution

$f (x) = {(x - 2)}^{- 5} + {(x - 2)}^{- 4} + 3 {(x - 2)}^{- 4} + 3 {(x - 2)}^{- 3} + 1 + {(x - 2)}^{8} + {(x - 2)}^{10} Say x-2 = t (> 0)$

$f (x) = t^{- 5} + t^{- 4} + 3 t^{- 3} + 1 + t^{8} + t^{10} . Now apply A M \geq G M$

$\frac{t^{- 5} + t^{- 4} + t^{- 3} + t^{- 3} + t^{- 3} + 1 + t^{8} + t^{10}}{8} \geq {(t^{- 5} . t^{- 4} . t^{- 3} . t^{- 3} .1. t^{8} . t^{10})}^{1 / 8} \Rightarrow f (x) \geq 8 \forall x > 2$