# Prove that the function $f(x)$ is increasing for $x<0$ and decreasing for $x>0$

I'm interested by the following problem :

Let $$a,b,c>0$$ and $$x>0$$ then the following function is decreasing : $$f(x)=\sqrt{\frac{(abc)^x}{a^x+b^x+c^x}}\Big(\frac{1}{7a^x+b^x}+\frac{1}{7b^x+c^x}+\frac{1}{7c^x+a^x}\Big)$$ Furthermore for $$x<0$$ the function above is increasing .

Try : If we derive we get : $$f'(x)=\sqrt{\frac{(a b c)^x}{a^x + b^x + c^x}} \Big(-\frac{(7 a^x \log(a) + b^x \log(b))}{(7 a^x + b^x)^2} - \frac{(a^x \log(a) + 7 c^x \log(c))}{(a^x + 7 c^x)^2} - \frac{(7 b^x \log(b) + c^x \log(c))}{(7 b^x + c^x)^2}\Big) +\frac{ \Big(\frac{1}{7a^x+b^x}+\frac{1}{7b^x+c^x}+\frac{1}{7c^x+a^x}\Big) \Big(\frac{(a b c)^x \log(a b c)}{(a^x + b^x + c^x)} - (a b c)^x \frac{(a^x \log(a) + b^x \log(b) + c^x \log(c))}{(a^x + b^x + c^x)^2}\Big)}{2 \sqrt{\frac{(a b c)^x}{(a^x + b^x + c^x)}}}$$

If we multiply by $$x$$ and make a little substitution $$u=a^x$$ we have : $$xf'(x)=\sqrt{\frac{(uvw)}{u+v+w}} \Big(-\frac{(7 u \log(u) +v \log(v))}{(7 u + v)^2} - \frac{(u \log(u) + 7 w \log(w))}{(u + 7 w)^2} - \frac{(7 v \log(v) + w \log(w))}{(7 v + w)^2}\Big) +\frac{ \Big(\frac{1}{7u+v}+\frac{1}{7v+w}+\frac{1}{7w+u}\Big) \Big(\frac{(uvw) \log(uvw)}{(u+v+w)} - (uvw) \frac{(u \log(u) + v \log(v) + w \log(w))}{(u + v + w)^2}\Big)}{2 \sqrt{\frac{(uvw)}{(u+v+w)}}}$$

But this last inequality is too monstruous for me so if you have another way it would be nice .

• You want to show that $$\frac{\sum\limits_{\text{cyc}}a^x\ln a}{a^x+b^x+c^x}+2\sum\limits_{\text{cyc}}\frac{7a^x\ln a+b^x\ln b}{(7a^x+b^x)^2}>\ln abc$$ for $x>0$. – TheSimpliFire Jun 8 at 15:42