# How find the minimum $\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$,if $x,y,z>0$

let $x,y,z>0$, find the minimum of the value $$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$

I think we can use AM-GM inequality to find it.

$$5y+2=y+y+y+y+y+1+1\ge 7\sqrt[7]{y^5}$$ $$2x+5=x+x+1+1+1+1+1\ge 7\sqrt[7]{x^2}$$ $$x+3y=x+y+y+y\ge 4\sqrt[4]{xy^3}$$ $$3x+z=x+x+x+z\ge 4\sqrt[4]{x^3z}$$ but this is not true,because not all four equalities can hold at once.

This problem is from china middle school test,so I think have without Lagrange methods,so I think this inequality have AM-GM inequality

## 5 Answers

Here is another way, using Holder's inequality: $$\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz} \geqslant \frac{(\sqrt[4]{5y \cdot 5\cdot x\cdot z}+\sqrt[4]{2\cdot2z\cdot3y\cdot3x})^4}{xyz} \\ = (\sqrt5+\sqrt6)^4=241+44\sqrt{30}$$

with equality iff $\dfrac{5y}2=\dfrac5{2z}=\dfrac{x}{3y}=\dfrac{z}{3x} \iff (x, y, z) = \left(1, \sqrt\frac2{15}, \sqrt\frac{15}2 \right)$.

Here is yet another way, using (weighted) AM-GM. I would still prefer Holder (the solution posted earlier) for its simplicity, this is just to illustrate it can be done - especially if you know the point of equality.

Let $a = \sqrt{\frac2{15}}$. Then using weighted AM-GM, we can write the following inequalities. Note that they are constructed so that the point of equality is easily maintained: $$5y+2=5a \cdot \frac{y}a+2 \cdot 1 \ge (5a+2)\left(\frac ya \right)^{\frac{5a}{5a+2}}$$ $$\parallel ly, \quad 2z+5=\frac2a \cdot az+5\cdot1 \ge (\tfrac2a+5)(az)^{\frac2{5a+2}}$$ $$x+3y = 1\cdot x + 3a\cdot\frac{y}a \ge (1+3a)x^{\frac1{3a+1}}\left(\frac ya\right)^{\frac{3a}{3a+1}}$$ $$3x+z = 3\cdot x + \frac1a \cdot az \ge (3+\tfrac1a)x^{\frac{3a}{3a+1}}(az)^{\frac1{3a+1}}$$

Multiplying the lot, and substituting the value of $a$, we get $$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz} \ge 241+44\sqrt{30}$$

Let $$f(x,y,z)=\log(5y+2)+\log(2z+5)+\log(x+3y)+\log(3x+z)-\log(xyz).$$ Then $$f_x=\frac{-1}{x}+\frac{1}{x+3y}+\frac{3}{3x+z};\\ f_y=\frac{-1}{y}+\frac{3}{x+3y}+\frac{5}{2+5y};\\ f_z=\frac{-1}{z}+\frac{1}{3x+z}+\frac{2}{5+2z}.$$ So the FOC's yield $$x=1,\quad y=\sqrt{2/15}, \quad z = \sqrt{15/2}.$$ With these values, the original function evaluates to $241+44\sqrt{30}$. To be rigorous, you can verify the SOC's but I think it will probably work. According to Mathematica's FindMinimum, the values of $x$, $y$, and $z$ above solve the problem.

P.s. A commment on why naive AM-GM may not work: Suppose you split the 4 expressions according to \begin{gather} \frac{x}{m}+\cdots+\frac{x}{m}+\frac{3y}{n}+\cdots+\frac{3y}{n}\tag{i}\\ \frac{3x}{p}+\cdots+\frac{3x}{p}+\frac{z}{q}+\cdots+\frac{z}{q},\tag{ii}\\ \frac{5y}{r}+\cdots+\frac{5y}{r}+\frac{2}{s}+\cdots+\frac{2}{s},\tag{iii}\\ \frac{2z}{c}+\cdots+\frac{2z}{c}+\frac{5}{d}+\cdots+\frac{5}{d}.\tag{iv} \end{gather} The first 2 equations say $$x=\frac{3m}{n}y,\quad z=\frac{3q}{p}x.$$ You want the exponent of $x$ after applying AM-GM to match the exponent of $x$ in the denominator, which is $1$, so $\frac{m}{m+n}+\frac{p}{p+q}=1$, which simplifies to $\frac{m}{n}=\frac{q}{p}$. So (i) and (ii) imply $$z=\frac{9m^2}{n^2}y.\tag{A}$$ Next, (iii) implies $y=\frac{2r}{5s}$. And since we want $\frac{n}{n+m}+\frac{r}{r+s}=1$, it must be that $\frac{r}{s}=\frac{m}{n}$. So (iii) implies $$y=\frac{2}{5}\frac{m}{n}.\tag{B}$$ Similarly, (iv) implies $z=\frac{5c}{2d}$ with $\frac{c}{d}=\frac{p}{q}$ due to $\frac{c}{c+d}+\frac{q}{p+q}=1$. But $\frac{p}{q}=\frac{n}{m}$, so (iv) implies $$z=\frac{5}{2}\frac{n}{m}.\tag{C}$$ Putting together (A), (B), and (C), we see that $$9\frac{m^2}{n^2}=\frac{z}{y}=\frac{5n}{2m}/\frac{2m}{5n}\implies\frac{9m^2}{n^2}=\frac{25n^2}{4m^2}.$$ Taking square roots and simplifying yield $$6m^2=5n^2.$$ But now we have a problem because $\sqrt{\frac{5}{6}}$ cannot be rational.

$$f(x,y,z)=\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$ $$\nabla f= \frac{(3 (2+5 y) (5+2 z) (x^2-y z))}{(x^2 y z)} \hat e_x+\frac{(-((2 x-15 y^2) (3 x+z) (5+2 z))}{(x y^2 z))}\hat e_y+\frac{(-((x+3 y) (2+5 y) (15 x-2 z^2))}{(x y z^2))}\hat e_z$$ We get 4 real solutions: $$(5/6,-5/18,-5/2),(1,-\sqrt{2/15},\sqrt{15/2}),(1,\sqrt{2/15},\sqrt{15/2}),(6/5,-2/5,-18/5)$$ Out of which only $(1,\sqrt{2/15},\sqrt{15/2})$ is acceptable due to $x,y,z\ge0$ Now the value of function at this point is: $$f(1,\sqrt{2/15},\sqrt{15/2})=241+44\sqrt{30}$$

• Heh, we had a similar idea. Cheers. :) Commented Aug 30, 2014 at 7:33
• @KimJongUn :):) Commented Aug 30, 2014 at 7:39
• great solution!
– mike
Commented Aug 30, 2014 at 8:10

Hint:

$5y+2 \geq 2\sqrt{10y}$

$2z+5 \geq 2\sqrt{10z}$

$x+3y \geq 2\sqrt{3xy}$

$3x+z \geq 2\sqrt{3xz}$

• there's no guarantee that this gives the minimum... not all four equalities can hold at once. Commented Aug 30, 2014 at 5:56
• It looks like the hint is correct!
– mike
Commented Aug 30, 2014 at 6:56
• @mike the first inequality holds when $y=2/5$, second when $z=5/2$, third when $x=3y=6/5$ and fourth when $z=3x=18/5$ contradiction Commented Aug 30, 2014 at 7:01
• Close, but not quite. This shows $LHS \ge 480$, which is correct as the minimum is $\approx 481.998$, but equality is never possible, so this does not yield the min. Commented Aug 30, 2014 at 12:04