Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$

Question

So I am trtying to proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$. First I tried with Cauchy–Schwarz inequality but got nowhere.
Now I am trying to find a convex function, so I can use jensen's inequality, but I can't come up with one which works.. Has anyone an idea?

Answer 1

Another way. Let $a+b+c=3$.

Hence, we need to prove that $$\sum\limits_{cyc}\frac{a}{9-2a}\geq\frac{3}{7}$$ or $$\sum\limits_{cyc}\left(\frac{a}{9-2a}-\frac{1}{7}\right)\geq0$$ or $$\sum\limits_{cyc}\frac{a-1}{9-2a}\geq0$$ or $$\sum\limits_{cyc}\left(\frac{a-1}{9-2a}-\frac{a-1}{7}\right)\geq0$$ or $$\sum\limits_{cyc}\frac{(a-1)^2}{9-2a}\geq0.$$ Done!

Answer 2

Note that \begin{align*} (a+b+c)^2 \ge 3ab+3bc+3ac. \end{align*} Therefore, \begin{align*} &\ \frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b}\\ =&\ \frac{a^2}{a^2+3ab+3ac}+\frac{b^2}{b^2+3ab+3bc}+\frac{c^2}{c^2+3ac+3bc}\\ \ge&\ \frac{(a+b+c)^2}{a^2+b^2+c^2 + 6ab + 6ac+6bc}\\ =&\ \frac{(a+b+c)^2}{(a+b+c)^2 + 4ab + 4ac+4bc}\\ \ge&\ \frac{(a+b+c)^2}{(a+b+c)^2 + \frac{4}{3}(a+b+c)^2}\\ =&\ \frac{3}{7}. \end{align*}

Answer 3

Cauchy-Schwarz actually works pretty well.

We have with Cauchy-Schwarz if we put $x_1=\sqrt{\frac{a}{a+3b+3c}}$, $x_2=\sqrt{\frac{b}{b+3c+3a}}$ and $x_3=\sqrt{\frac{c}{c+3a+3b}}$ as well as $y_1=\sqrt{a(a+3b+3c)}$, $y_2=\sqrt{b(b+3c+3a)}$ and $y_3=\sqrt{c(c+3a+3b)}$ $$ \left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2+y_2^2+y_3^2\right)≥\left(x_1y_1+x_2y_2+x_3y_3\right)^2\iff\\ \left(\frac{a}{a+3b+3c}+\frac{b}{b+3c+3a}+\frac{c}{c+3a+3b}\right)\left(a(a+3b+3c)+b(b+3c+3a)+c(c+3a+3b)\right)≥(a+b+c)^2 $$ so its enough to prove $$ \frac{(a+b+c)^2}{\sum_{cyc}a(a+3b+3c)}≥\frac37\iff\\ 7(a^2+b^2+c^2+2ab+2bc+2ca)≥3(a^2+b^2+c^2+6ab+6bc+6ca)\iff\\ a^2+b^2+c^2≥ab+bc+ca $$ which is true.

Answer 4

Hint

The inequality is homogeneous in $(a,b,c)$ so we can choose them such that

$$a+b+c=\frac{1}{3}$$

so we have to find the minimum for the function:

$$f(a,b,c)=\frac{a}{1-2a}+\frac{b}{1-2b}+\frac{c}{1-2c}$$

with boundary

$$a+b+c=\frac{1}{3}$$

Now can use Lagrange Multiplier and find the minimum $3/7$

Answer 5

moving $$\frac{3}{7}$$ to the left and clearing the denominators we get $$3\,{a}^{3}-{a}^{2}b-{a}^{2}c-a{b}^{2}-3\,abc-a{c}^{2}+3\,{b}^{3}-{b}^{ 2}c-b{c}^{2}+3\,{c}^{3} \geq 0$$ and this is equivalent to $$a^3+b^3+c^3-3abc+(a-b)(a^2-b^2)+(b^2-c^2)(b-c)+(a^2-c^2)(a-c)\geq 0$$ which is clear (the first Terms are AM-GM)

Answer 6

WLOG $a\geq b\geq c.$ $$\text {Let } x=1/(a+3b+3c),\; y=1/(b+3c+3a),\; z=1/(c+3a+3b).$$ We have $a+3b+3c\leq b+3c+3a\leq c+3b+3a.$ Therefore $$(1).\quad x\geq y\geq z>0.$$ Let $f(a,b,c)=ax+by+cz.$

Differentiating $f$ by $c,$ keeping $a$ and $b$ constant, we have $$\partial f/\partial c=-3ax^2-3by^2+3(a+b)z^2=3(a(z^2-x^2)+b(z^2-y^2)).$$ This is not positive by (1). Therefore $$(2).\quad f(a,b,c)\geq f(a,b,b).$$ Differentiating $f$ by $a,$ keeping $b$ and $c$ constant, we have $$\partial f/\partial a=(3b+3c)x^2-3by^2-3cz^2=3(b(x^2-y^2)+c(x^2-y^2)).$$ This is not negative by (1). Therefore $$(3). \quad f(a,b,c)\geq f(b,b,c).$$ By(2) and (3) we have $$f(a,b,c)\geq f(b,b,c)\geq f(b,b,b)=3/7.$$

Answer 7

We need to prove that $$\sum\limits_{cyc}\left(\frac{a}{a+3b+3c}-\frac{1}{7}\right)\geq0$$ or $$\sum\limits_{cyc}\frac{2a-b-c}{a+3b+3c}\geq0$$ or $$\sum\limits_{cyc}\frac{a-b-(c-a)}{a+3b+3c}\geq0$$ or $$\sum_{cyc}(a-b)\left(\frac{1}{a+3b+3c}-\frac{1}{b+3c+3a}\right)\geq0$$ or $$\sum\limits_{cyc}\frac{(a-b)^2}{(a+3b+3c)(b+3c+3a)}\geq0.$$ Done!