# Problem in Arithmetic Mean - Geometric Mean inequality [duplicate]

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

• @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.
– dxiv
Jun 30, 2018 at 1:11
• @Ehit Karim $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{24\sqrt[3]{abc}}{a+b+c} \geq 11$ is also true. Jun 30, 2018 at 3:27

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.