# Prove that $({a\over a+b})^3+({b\over b+c})^3+ ({c\over c+a})^3\geq {3\over 8}$

Let $$a,b,c$$ be positive real numbers. Prove that $$\Big({a\over a+b}\Big)^3+\Big({b\over b+c}\Big)^3+ \Big({c\over c+a}\Big)^3\geq {3\over 8}$$

If we put $$x=b/a$$, $$y= c/b$$ and $$z=a/c$$ we get $$xyz=1$$ and

$$\Big({1\over 1+x}\Big)^3+\Big({1\over 1+y}\Big)^3+ \Big({1\over 1+z}\Big)^3\geq {3\over 8}$$

Since $$f(x)=\Big({1\over 1+x}\Big)^3$$ is convex we get, by Jensen,: $$\Big({1\over 1+x}\Big)^3+\Big({1\over 1+y}\Big)^3+ \Big({1\over 1+z}\Big)^3\geq 3f({x+y+z\over 3})$$

Unfortunately, since $$f$$ is decreasing we don't have $$f({x+y+z\over 3}) \geq f(1) = {1\over 8}$$.

Some idea how to solve this?

• How is this question "seeking personal advice," whoever voted to close this thread? – Batominovski Dec 4 '18 at 21:26

Let $$g(t):=\left(\dfrac{1}{1+\exp(t)}\right)^3$$ for $$t\in\mathbb{R}$$. Then, $$g''(t)=\frac{3\,\exp(t)\,\big(3\,\exp(t)-1\big)}{\big(1+\exp(t)\big)^5}\text{ for each }t\in\mathbb{R}\,.$$ Thus, $$g$$ is convex on $$\big[-\ln(3),\infty\big)$$.

Let $$x:=\dfrac{b}{a}$$, $$y:=\dfrac{c}{b}$$, and $$z:=\dfrac{a}{c}$$ be as the OP defines. Then, the required inequality is equivalent to $$g\big(\ln(x)\big)+g\big(\ln(y)\big)+g\big(\ln(z)\big)\geq \dfrac{3}{8}\,.\tag{*}$$ If $$x$$, $$y$$, or $$z$$ is less than $$\dfrac{1}{3}$$, then clearly the left-hand side of (*) is greater than $$\left(\dfrac{1}{1+\frac13}\right)^3=\frac{27}{64}>\frac38\,.$$ If all $$x$$, $$y$$, and $$z$$ are greater than or equal to $$\dfrac13$$, then $$\ln(x),\ln(y),\ln(z)\geq -\ln(3)$$, so that we can use convexity of $$g$$ on $$\big[-\ln(3),\infty\big)$$. By Jensen's Inequality, $$g\big(\ln(x)\big)+g\big(\ln(y)\big)+g\big(\ln(z)\big)\geq3\,g\left(\frac{\ln(x)+\ln(y)+\ln(z)}{3}\right)=3\,g(0)=\frac{3}{8}\,.$$ Hence, the equality holds if and only if $$x=y=z=1$$, making $$a=b=c$$.

• I don't get it. What is the difference with my solution? – Aqua Dec 4 '18 at 21:45
• I'm not sure what to say to that, but my idea was to bypass the inequality $\frac{x+y+z}{3}\geq \sqrt[3]{xyz}$ that you would need to use in your attempt. – Batominovski Dec 4 '18 at 21:48

This is more of a comment, but I don't have the reputation. Use Lagrange multipliers. Solving, you find that the critical points occur when $$xyz=1$$ and $$yz (1+x)^4 = xz (1+y)^4 = xy (1+z)^4$$. I think the only solution is $$x=y=z=1$$. Clearly, it's a minimum and plugging in shows the bound.

We can rewrite the condition as $$xyz=1$$ and $$\frac{(1+x)^4}{x} = \frac{(1+y)^4}{y} = \frac{(1+z)^4}{z}$$

The function $$g(x)=\frac{(1+x)^4}{x}$$ is decreasing from $$0$$ to $$1/3$$ and increasing from $$1/3$$ to $$\infty$$. This shows that 2 of $$x,y,z$$ must be equal (WLOG $$x$$ and $$y$$) and $$z=1/x^2$$. It remains to solve $$\frac{(1+x)^4}{x} = \frac{(1+1/x^2)^4}{1/x^2}$$. This time, it's not hard to check $$x=1$$ is the only solution and we are done.

• It seems to me there's enough here to post as an answer, no need to apologize. – David K Dec 4 '18 at 21:11

By Holder $$\left(\sum_{cyc}\frac{a^3}{(a+b)^3}\right)^2\sum_{cyc}1\geq\left(\sum_{cyc}\sqrt[3]{\left(\frac{a^3}{(a+b)^3}\right)^2\cdot1}\right)^3=\left(\sum_{cyc}\frac{a^2}{(a+b)^2}\right)^3.$$ Thus, it's enough to prove that $$\frac{\left(\sum\limits_{cyc}\frac{a^2}{(a+b)^2}\right)^3}{3}\geq\frac{9}{64}$$ or $$\sum\limits_{cyc}\frac{a^2}{(a+b)^2}\geq\frac{3}{4}.$$ Now, by C-S $$\sum\limits_{cyc}\frac{a^2}{(a+b)^2}=\sum\limits_{cyc}\frac{a^2(a+c)^2}{(a+b)^2(a+c)^2}\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}(a+b)^2(a+c)^2}.$$ Thus, it's enough to prove that $$4\left(\sum\limits_{cyc}(a^2+ab)\right)^2\geq3\sum\limits_{cyc}(a+b)^2(a+c)^2,$$ which is true even for all reals $$a$$, $$b$$ and $$c$$.

Indeed, the last inequality is symmetric inequality by degree four,

which says that by $$uvw$$ (https://math.stackexchange.com/tags/uvw/info )

it's enough to prove the last inequality for equality case of two variables and since

it's the homogeneous inequality by even degree, we can assume $$b=c=1$$, which gives $$(a-1)^2(a+3)^2\geq0.$$ Done!