An exponential equation over positive real numbers $4^x+14^x+3^x=11^x+10^x$ Solve the following equation over the positive reals:
$$4^x+14^x+3^x=11^x+10^x.$$
By inspecting the graph, the solutions must be $x \in \{1,2\}$
I tried using inequalities like $3^x+4^x<5^x$ for $x>2$ but I couldn’t work it forward. Thanks in advance!

We'll solve this equation for any real value of $x$.
Indeed, let $x>2$ and $x=1+t$.
Thus, $t>1$ and  since $f(x)=x^t$ is a convex function, by Jensen we obtain:
$$4^x+3^x+14^x=(3\cdot3^t+8\cdot14^t)+\left(4\cdot4^t+6\cdot14^t\right)\geq$$
$$\geq11\left(\frac{3\cdot3+8\cdot14}{11}\right)^t+10\left(\frac{4\cdot4+6\cdot14}{10}\right)^t=11^{t+1}+10^{t+1}=11^x+10^x.$$
The equality does not occur, which says that our equation has no roots for $x>2$.
Similarly, for $x<1$ let $x=1+t,$ where $t<0$ and since $g(x)=x^t$ is a convex function,
by the same Jensen we obtain that our equation has no roots for $x<1$.
Now, let $1\leq x\leq 2$ and $$h(x)=\left(\frac{3}{11}\right)^x+\left(\frac{4}{11}\right)^x+\left(\frac{14}{11}\right)^x-\left(\frac{10}{11}\right)^x-1$$ and we'll prove that $h$ is a convex function.
Indeed, $$h''(x)=\left(\frac{3}{11}\right)^x\ln^2\frac{11}{3}+\left(\frac{4}{11}\right)^x\ln^2\frac{11}{4}+\left(\frac{14}{11}\right)^x\ln^2\frac{14}{11}-\left(\frac{10}{11}\right)^x\ln^2\frac{11}{10}\geq$$
$$\geq\left(\tfrac{3}{11}\right)^2\ln^2\tfrac{11}{3}+\left(\tfrac{4}{11}\right)^2\ln^2\tfrac{11}{4}+\tfrac{14}{11}\ln^2\tfrac{14}{11}-\tfrac{10}{11}\ln^2\tfrac{11}{10}>\tfrac{10}{11}\ln^2\tfrac{14}{11}-\tfrac{10}{11}\ln^2\tfrac{11}{10}>0,$$
which says that our equation has maximum two roots on $[1,2].$ 
But $1$ and $2$ are roots and we are done!
