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Let a,b,c be positive real numbers, prove that $$ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$$

I am suppose to use AM-GM inequality, I tried $$ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3 $$ and $$ a + b + c\geq 3 \sqrt[3]{abc} $$ implying $$ \frac{3\sqrt[3]{abc}}{a+b+c} \leq 1 $$ Now adding the two inequality could give me the desired result but the problem I face is with second inequality sign(it's less than or equal to 1 rather than greater

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marked as duplicate by mechanodroid, dxiv, Xander Henderson, Michael Rozenberg inequality Jun 30 '18 at 4:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ @mechanodroid There is a certain irony in that the (wrongly) accepted answer there makes exactly the error that the OP (correctly) caught in their question here. $\endgroup$ – dxiv Jun 30 '18 at 1:11
  • $\begingroup$ @Ehit Karim $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{24\sqrt[3]{abc}}{a+b+c} \geq 11$ is also true. $\endgroup$ – Michael Rozenberg Jun 30 '18 at 3:27
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I will go the standard way. Let us introduce $A,B,C>0$ with $$ a=A^3\ ,\ b=B^3\ ,\ c=C^3\ . $$ Then we have to show $$ \frac{A^3}{B^3}+ \frac{B^3}{C^3}+ \frac{C^3}{A^3}+ \frac{3ABC}{A^3+B^3+C^3} -4\ge 0 \ . $$ We multiply with $A^3B^3C^3(A^3+B^3+C^3)$, and have to show equivalently: $$ \begin{aligned} &A^9C^3+B^9A^3+C^9B^3\\ &\qquad+A^6B^6+B^6C^6+C^6A^6\\ &\qquad\qquad+3A^4B^4C^4\\ &\qquad\qquad\qquad\qquad\qquad\qquad\ge 3A^6B^3C^3 + 3A^3B^6C^3 +3A^3B^3C^6 \ . \end{aligned} $$ Let us see how to dominate the terms on the R.H.S of $\ge $ with the ones on the $L.H.S$. Consider first $3A^6B^3C^3$. The degree is $(6,3,3)$. We consider it in the plane $X+Y+Z=12$, and search for a triangle using the weights

  • $(9,0,3)$, $(3,9,0)$, $(0,3,9)$,
  • $(6,6,0)$, $(6,0,6)$, $(0,6,6)$,
  • $(4,4,4)$,

in the same plane, which contains $(6,3,3)$ in its interior. We do so, since we want to apply the inequality $$ a_1x_1+a_2x_2+a_3x_3+\dots\ge x_1^{a_1}\cdot x_2^{a_2}\cdot x_3^{a_3}\cdot\dots \ , $$ where $a_1,a_2,a_3,\dots$ are positive weights.

  • From $\frac 47(9,0,3)+\frac 27(3,9,0)+\frac 17(0,3,9)=(6,3,3)$ we obtain the inequality: $$ \frac 47A^9C^3 + \frac 27A^3B^9 + \frac 17B^3C^9 \ge (A^9C^3)^{4/7}\cdot (A^3B^9)^{2/7}\cdot (B^3C^9)^{1/7} = A^6B^3C^3\ . $$

  • Cyclically doing this, we can "cover" (dominate) once the sum $A^6B^3C^3 + A^3B^6C^3 +A^3B^3C^6 $.

  • But the problem with this domination is that we have not used our weakest term, the one with $(4,4,4)$, but instead we have lost the strongest. So we have to combine. A general combination would be of the shape: $$ \left(\frac 47-\frac r3\right) (9,0,3) + \left(\frac 27-\frac r3\right) (3,9,0) + \left(\frac 17-\frac r3\right) (0,3,9) +r(4,4,4) =(6,3,3) \ . $$ Then best value we can use for $r$ is $r=\frac 37$. This gives $$ \frac 37 (9,0,3) + \frac 17 (3,9,0) + \frac 37(4,4,4) =(6,3,3) \ . $$

  • This gives then an inequality of the same shape as above, we multiply it with $\frac 74$ to get: $$ \frac 34A^9C^3 + \frac 14A^3B^9 + \frac 34A^4B^4C^4 \ge \frac 74A^6B^3C^3\ . $$

  • Cyclically doing this, we can "cover" (dominate) $7/4$ of the sum $A^6B^3C^3 + A^3B^6C^3 +A^3B^3C^6 $.

  • For the rest, we know what we still have to show, this is $$ \begin{aligned} &A^6B^6+B^6C^6+C^6A^6\\ &\qquad+\frac34A^4B^4C^4\\ &\qquad\qquad\qquad\qquad\ge \frac 54(A^6B^3C^3 + A^3B^6C^3 +A^3B^3C^6) \ . \end{aligned} $$

  • We multiply with $\frac 45$, have to show equivalently $$ \frac 45A^6B^6+\frac 45B^6C^6+\frac 45C^6A^6+\frac35A^4B^4C^4 \ge A^6B^3C^3 + A^3B^6C^3 +A^3B^3C^6\ . $$

  • This follows from $$ \frac 25A^6B^6+\frac 25A^6C^6+\frac15A^4B^4C^4 \ge A^6B^3C^3\ , $$

(after we cycle and add). The last inequality corresponds to $$ \frac 25(6,6,0)+\frac 25(6,0,6)+\frac 15(4,4,4)=(6,3,3)\ . $$


Note: This proof can be now rewritten to fit in a few lines, however, i prefer to give it all... There was no point where i did something unnatural, so this is a natural solution.

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