Let $a$, $b$, $c$, $x$, $y$ and $z$ be positives such that $a+b+c=x+y+z=1$. Prove that: $$\frac{a}{y+z}+\frac{b}{x+z}+\frac{c}{x+y}\geq1+\frac{27}{2}abc$$

I tried to make the homogenization:

We need to prove that $$(x+y+z)\left(\frac{a}{y+z}+\frac{b}{x+z}+\frac{c}{x+y}\right)\geq1+\frac{27}{2}abc$$ or $$\frac{ax}{y+z}+\frac{by}{x+z}+\frac{cz}{x+y}\geq\frac{27abc}{2(a+b+c)^2}$$ and what is the rest?

Maybe now we need to get rid of $x$, $y$ and $z$, but I don't see how we can do it.

Thank you!


I used Lagrange Multipliers here. I'd love to see a way that doesn't rely on them, because I feel there should be a much snappier solution. I'm a bit rusty with contest math, but I think I've checked my own work well enough. That's no excuse for holes, but I offer it as a caveat. It's likely there can be improvements in the style, etc., even though I've checked several times for holes. If people have improvements, or point out outright errors, I'm of course open to that.

Let's look at the left hand side of the second equation you offer and try to see what values of x/y/z maximize it.

$\frac{ax}{1-x} + \frac{by}{1-y} + \frac{cz}{1-z} \ge \frac{27abc}{2(a+b+c)^2}$ (a)

The left hand side can be expressed as

$\frac{a}{1-x} + \frac{b}{1-y} + \frac{c}{1-z} - 1$

(adding a, b and c to the respective fractions)

and thus has a Lagrange function of

$\frac{a}{1-x} + \frac{b}{1-y} + \frac{c}{1-z} - 1 - \lambda(x+y+z)$

If we differentiate with respect to X, we get

$\lambda = \frac{a}{x^2}$

With respect to y, $\lambda = \frac{b}{y^2}$ and with z, $\lambda = \frac{c}{z^2}$.

That means $\frac{x}{\sqrt{a}} = \frac{y}{\sqrt{b}}$ or $y = x\sqrt{b/a}$. Similarly $z = x\sqrt{c/a}$.

Since x+y+z=1, then, $x(\sqrt{a}+\sqrt{b}+\sqrt{c})/\sqrt{a} = 1$, or $x = \sqrt{a}/(\sqrt{a}+\sqrt{b}+\sqrt{c})$. Similarly $y = \sqrt{b}/(\sqrt{a}+\sqrt{b}+\sqrt{c})$ and $z = \sqrt{c}/(\sqrt{a}+\sqrt{b}+\sqrt{c})$.

Now we can express (a) as follows, noting $\frac{x}{1-x} = \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}}$.

$\frac{a\sqrt{a}}{\sqrt{b}+\sqrt{c}} + \frac{b\sqrt{b}}{\sqrt{a}+\sqrt{c}} + \frac{c\sqrt{c}}{\sqrt{b}+\sqrt{a}} \ge \frac{27abc}{(a+b+c)^2}$

Now here's where I think there must be a better solution, but I multiplied everything out by $((a+b+c)^2)(\sqrt{b}+\sqrt{a})(\sqrt{a}+\sqrt{c})(\sqrt{b}+\sqrt{c})$ and compared the polynomial coefficients it worked okay. The strategy here is to look at the coefficients $\sqrt{abc}$ on the left and right side with the AM-GM inequality, then to deal with the symmetrical $\sqrt{a}+$\sqrt{b}+$\sqrt{c}$.

Let's look at the right hand side, since it's simpler to expand.

$(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})(\sqrt{b}+\sqrt{c})$ becomes $a\sqrt{b} + a\sqrt{c} + b\sqrt{a} + b\sqrt{c} + c\sqrt{a} + c\sqrt{b} + 2\sqrt{abc}$ and that means:

The right side's $\sqrt{abc}$ term is $27abc$, and its $\sqrt{a}$ term is $27abc(b+c)/2$.

The left side is, of course, trickier.

$(a+b+c)^2 (a\sqrt{a}(\sqrt{a}+\sqrt{c})(\sqrt{a}+\sqrt{b})+b\sqrt{b}(\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b})+c\sqrt{c}(\sqrt{a}+\sqrt{c})(\sqrt{c}+\sqrt{b}))$

Let's isolate the abc-term, without expanding $(a+b+c)^2$. Each of the three products has one way to get a term with $\sqrt{abc}$ so that gives

$(a+b+c)^2 (a\sqrt{abc}+b\sqrt{abc}+c\sqrt{abc})$ which simplifies to

$(a+b+c)^3\sqrt{abc}$ but $a+b+c/3 > \sqrt[3]{abc}$ by the AM GM inequality, and cubing both sides of that equation gives

$(a+b+c)^3 > 27abc$. In other words, the $\sqrt{abc}$ term on the left of the equation is greater than or equal to the $\sqrt{abc}$ term on the right.

Now let's look at the left, ignoring the $\sqrt{abc}$ coefficients. Let's get rid of the $(a+b+c)^2$ for the moment.

$a\sqrt{a}(a+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}) + b\sqrt{b}(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}) + c\sqrt{c}(c+\sqrt{bc}+\sqrt{ac}+\sqrt{ab})$

$(a^2+b^2+c^2)\sqrt{a} + (a^2+b^2+c^2)\sqrt{b} + (a^2+b^2+c^2)\sqrt{c}$

So the left's remaining coefficients are $(a+b+c)^2(a^2+b^2+c^2)(\sqrt{a}+\sqrt{b}+\sqrt{c})$.

Let's look at the right. It expands to

$27abc/2 * (a(\sqrt{b}+\sqrt{c}) + b(\sqrt{a}+\sqrt{c}) + c(\sqrt{b}+\sqrt{a}))$

Let's use the rearrangement inequality here. The sum in parentheses is the smallest for $x_1=a,x_2=b,x_3=c$ and $y_1=\sqrt{b}+\sqrt{c}, y_2=\sqrt{a}+\sqrt{c},y_3=\sqrt{b}+\sqrt{a})$.

If we let $z_1 = y_1 - (\sqrt{a}+\sqrt{b}+\sqrt{c})$ then we notice y_j decreases if we increase x_j, so we have the case where we have the rearrangement of the lowest sum. Thus, it is less than or equal to the average of all the possible sums, in other words,

$(a(\sqrt{b}+\sqrt{c}) + b(\sqrt{a}+\sqrt{c}) + c(\sqrt{b}+\sqrt{a})) < (a+b+c) * (\sqrt{a}+\sqrt{b}+\sqrt{c})*2/3$

In other words, the right is less than $9abc * (a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c})$

That puts us in the home stretch. We need to show

$(a+b+c)^2(a^2+b^2+c^2)(\sqrt{a}+\sqrt{b}+\sqrt{c}) > 9abc(a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c})$

But a lot cancels out.

$(a+b+c)(a^2+b^2+c^2) > 9abc$

However, $a+b+c>3\sqrt[3]{abc}$ and $(a^2+b^2+c^2) > 3\sqrt[3]{abc}^2$ both by the AM GM inequalities.


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