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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?

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    $\begingroup$ How is this question "seeking personal advice," whoever voted to close this thread? $\endgroup$ Dec 4, 2018 at 21:26

3 Answers 3

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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 $a=b=c$.

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  • $\begingroup$ I don't get it. What is the difference with my solution? $\endgroup$
    – nonuser
    Dec 4, 2018 at 21:45
  • $\begingroup$ 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. $\endgroup$ Dec 4, 2018 at 21:48
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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.

Added:

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.

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    $\begingroup$ It seems to me there's enough here to post as an answer, no need to apologize. $\endgroup$
    – David K
    Dec 4, 2018 at 21:11
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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!

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