Prove that $a+b^2+c^3+d^4 \ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$

Asked Feb 07 '16 at 12:07

Active Feb 07 '16 at 16:25

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If $0 < a \le b \le c \le d$ and $abcd = 1$ prove that $$a+b^2+c^3+d^4 \ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$$

I first thought of multiplying both sides with $\big(\frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}\big)$

Then by applying C-S inequality in the LHS the inequality would be $$16 \ge \Big(\frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}\Big)^2$$

But if $(a,b,c,d)=(\frac14,1,1,4)$ the inequality does not hold true.

edited Feb 07 '16 at 16:25

wythagoras

asked Feb 07 '16 at 12:07

user302454

You can assume them to be $x,y..z$ and then use $AM>=HM$ – Archis Welankar Feb 07 '16 at 14:10
the RHS should have been $\frac{4}{\frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}}$ – user302454 Feb 07 '16 at 14:12
this is from Michael Rozenberg it is not so simple – Dr. Sonnhard Graubner Feb 07 '16 at 14:17
i have proved it by BW but this proof is ugly – Dr. Sonnhard Graubner Feb 07 '16 at 14:17

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