# Let a, b, c be positive real numbers. Prove that

Let a,b,c be positive real numbers. Prove that

$$\frac{a^3+b^3+c^3}{3}\geq\frac{a^2+bc}{b+c}\cdot\frac{b^2+ca}{c+a}\cdot\frac{c^2+ab}{a+b}\geq abc$$

I will post what I had solved originally, however it is unfortunately incorrect. Please help solve and/or aid in finding my mistakes :]

This is what I have although it is incorrect any help would be greatly appreciated.!

$\mathbf{1.)}$ Consider now the expression $\frac{a^3+b^3+c^3}{3}$

The expressions has minimum value given by:

$a\geq b \geq c > 0 \implies a^3 \geq b^3 \geq c^3$

Thus $\frac{a^3+b^3+c^3}{3}$ has minimum value $c^3$

$\mathbf{2.)}$ On the other hand, consider the expression $\frac{a^2+bc}{b+c}$

Again we have $a \geq b \geq c >0 \implies a^2+bc \geq c^2+c\cdot c = 2c^2$

$a \geq b \geq c > 0 \implies a^2+bc \leq a^2 + a \cdot a = 2a^2$

$a \geq b \geq c > 0 \implies b+c \geq 2c$

and $a \geq b \geq c > 0 \implies b+c \leq 2a$

$\mathbf{3.)}$ Finally we know that a positive fraction is maximized when the numerator is as large as possible and the denominator is as small as possible

Therefore we have $\frac{a^2+bc}{b+c} \leq \frac{2a^2}{2c} = \frac{a^2}{c} \leq a$

Using identical considerations, we have $\frac{b^2+ca}{ca} \leq \frac{a^2}{c} \leq a$ and $\frac{c^2+ab}{a+b}\leq \frac{a^2}{c} \leq a$

So that we have

Hence, $\frac{a^2+bc}{b+c} \cdot \frac{b^2+ca}{c+a} \cdot \frac{c^2+ab}{a+b} \leq a^3$

This is the maximum value of the expression

$\mathbf{4.)}$ The minimum value will be achieved for a fraction when the numerator is as small as possible and denominator is as large as possible

Again, by an overall similar consideration, we have $\frac{a^2+bc}{b+c} \geq \frac{a^2+c^2}{2b}$ is the minimum value of this expression

Similarly $\frac{b^2+ca}{c+a} \geq \frac{b^2+c^2}{2a}$ and $\frac{c^2+ab}{a+b} \geq \frac{c^2+b^2}{2a}$

Hence

But we have $a^2+b^2 \geq 2b^2$, $c^2+b^2 \geq 2c^2$

Hence,

Therefore, $\frac{a^2+bc}{b+c} \cdot \frac{b^2+ca}{c+a} \cdot \frac{c^2+ab}{a+b} \geq \frac{bc^4}{a^2}$

$\mathbf{5.)}$ And finally, we have $a^3 \geq abc$ is the maximum value of $abc$

The minimum value of $\frac{a^3+b^3+c^3}{3}$ is $c^3$ which is greater than $a^3$ which is the maximum value of $\frac{a^2+bc}{b+c}\cdot \frac{b^2+ca}{c+a} \cdot \frac{c^2+ab}{a+b}$

That is,

So that $\frac{a^3+b^3+c^3}{3} \geq \frac{a^2+bc}{b+c}\cdot \frac{b^2+ca}{c+a} \cdot \frac{c^2+ab}{a+b}$

Again we have $\frac{a^2+bc}{b+c}\cdot \frac{b^2+ca}{c+a} \cdot \frac{c^2+ab}{a+b} \geq \frac{bc^4}{a^2}$

• In the fourth section, the inequality $\frac{a^2 + bc}{b+c} > \frac{a^2+b^2} {2b}$ is incorrect. It should be $\frac{a^2 + bc}{b+c} > \frac{a^2+c^2} {2b}$ – Zack Ni Aug 16 '16 at 3:01
• Greatly appreciate all of the edits to allow for the ease of viewing, Can anyone comment if my thought process is somewhat correct ? Also any other known proofs that would apply to this inequality? – LKB Aug 16 '16 at 3:27
• No worries @LKB – bigfocalchord Aug 16 '16 at 3:29

The right inequality.

By AM-GM we obtain:

$$\prod\limits_{cyc}(a^2+bc)=\sqrt[3]{\frac{\prod\limits_{cyc}((a^2+bc)(b^2+ac))^2}{\prod\limits_{cyc}(a^2+bc)}}=\sqrt[3]{\frac{\prod\limits_{cyc}(ab(c^2+ab)+c(a^3+b^3))^2}{\prod\limits_{cyc}(a^2+bc)}}\geq$$ $$\geq\sqrt[3]{\frac{\prod\limits_{cyc}(4(a^3+b^3)(c^2+ab)abc)}{\prod\limits_{cyc}(a^2+bc)}}\geq\sqrt[3]{\frac{\prod\limits_{cyc}((a+b)^3(c^2+ab)abc)}{\prod\limits_{cyc}(a^2+bc)}}=abc\prod\limits_{cyc}(a+b)$$

The left inequality.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Since $\prod\limits_{cyc}(a^2+bc)=2a^2b^2c^2+\sum\limits_{cyc}(a^3b^3+a^4bc)=8w^6+A(u,v^2)w^3+B(u,v^2)$,

$\prod\limits_{cyc}(a+b)=9uv^2-w^3$ and $a^3+b^3+c^3=27u^3-27uv^2+3w^3$, we see that

our inequality is equivalent to $f(w^3)\geq0$, where $f$ is a concave function.

But the concave function gets a minimal value for an extremal value of $w^3$,

which happens for equality case of two variables and we must check $w^3\rightarrow0^+$.

1. Let $w^3\rightarrow0^+$. Let $c\rightarrow0^+$.

We get $(a^3+b^3)(a+b)ab\geq3a^3b^3$, which is obvious;

1. $b=c=1$, which gives $(a^2-1)^2(2a+1)\geq0$.

Done!

Also we can use a full expanding and we'll get something obvious:

$\sum\limits_{sym}\left(a^5b+a^4b^2-\frac{1}{2}a^4bc-\frac{3}{2}a^3b^3+a^3b^2c-a^2b^2c^2\right)\geq0$, which is Muirhead.

• For what is f is a concave function referring to ? I was asked to elaborate on a similar problem, but haven't had a chance to return to the problem – LKB Aug 16 '16 at 3:58
• @LKB $f(x)=-x^2+bx+c$ is a concave function. Draw it! If $k\leq x\leq m$ so $\min\limits_{x\in[k,m]}f=\min\{f(k),f(m)\}$. – Michael Rozenberg Aug 16 '16 at 4:02
• Thank you :] and will do ! – LKB Aug 16 '16 at 4:28
• I get the same symmetric sum, but I don't see how Muirhead makes that positive, because of all the different coefficients. – Mark Fischler May 2 '17 at 17:17
• @Mark Fischler By Muirhead $\frac{1}{2}\sum\limits_{cyc}(a^5b-a^4bc)\geq0$, $\frac{1}{2}\sum\limits_{cyc}(a^5b-a^3b^3)\geq0$, $\sum\limits_{cyc}(a^4b^2-a^3b^3)\geq0$ and $\sum\limits_{cyc}(a^3b^2c-a^2b^2c^2)\geq0$, – Michael Rozenberg May 2 '17 at 19:54