How to find the maximum value of $3^x + 5^x - 9^x + 15^x - 25^x$ as $x$ varies over the reals? 
How to find the maximum value of  $3^x + 5^x - 9^x + 15^x - 25^x$ as x
  varies over the reals ?

Suggestions please!
 A: Let $u=3^x$ and $v=5^x$. Then, $$3^x + 5^x - 9^x + 15^x - 25^x=u+v-u^2+uv-v^2=:f(u,v).$$
Taking the partial derivatives with respect to $u$ and $v$, $$f_u=1-2u+v,\quad f_v=1-2v+u.$$
Setting these values equal to $0$, $$v=2u-1, \quad v=\frac{u+1}{2}.$$ Solving these systems simultaneously, $$u=v=1 \implies 3^x=5^x=1 \implies x=0.$$
Therefore, the maximum occurs at $x=0$, with maximum value of 1.
EDIT: In order to show that this is indeeed a maximum value, we make use of the second derivative test for functions of two variables:
Theorem: Let the function $z = f (u, v)$ have continuous second order partial derivatives and let $(u_0,v_0)$ be a critical point of the function. Let $$D=f_{uu}(u_0,v_0)\cdot f_{vv}(u_0,v_0)-[f_{uv}(u_0,v_0)]^2.$$ If $D>0$ and $f_{uu}(u_0,v_0)<0$, then $f$ has a local maximum at $(u_0,v_0)$.
Computing the second order partial derivatives, we have: $$f_{uu}(u,v)=-2, \quad f_{vv}(u,v)=-2, \quad f_{uv}(u,v)=1.$$
Hence, $D=3>0$ and $f_{uu}(1,1)=-2<0$, and so we may conclude that the critical point is a maximum value.
A: If you don't want to use calculus: Substitute $u=3^x$ and $y=5^x$ as in JVV's answer, obtaining a paraboloid:
$$f(u,v)=u+v-u^2+uv-v^2=(u+v)-(u-v)^2-uv\,.$$
Get rid of the $uv$ term by substituting $z=(u+v)/2$ and $y=(u-v)/2$, obtaining
$$2z - (2y)^2 -(z+y)(z-y) = 1 - (z-1)^2 - 3y^2\,.$$
Plug back $u$, $v$ to get a clearer view of $f$'s behavior:
$$f(u,v) = 1 - \frac14(u+v-2)^2 - \frac34(u-v)^2
$$
From this we see that $f$ has a maximum value of $1$, attained when $u+v-2=0$ and $u-v=0$, i.e., $u=v=1$.
A: Almost Same as grant_chat.
Let $a=3^x$ and $b=5^x\;,$ Then expression convert into $f(a,b) = a+b-a^2+ab-b^2$
So $$f(a,b) = -\frac{1}{2}[2a^2+2b^2-2ab-2a-2b]$$
So $$f(a,b) = -\frac{1}{2}\left[(a^2-2a+1)+(b^2-2b+1)+(a^2+b^2-2ab)-2\right]$$
So $$f(a,b) = -\frac{1}{2}\left[(a-1)^2+(b-1)^2+(a-b)^2-2\right]$$
So $$f(a,b) = \frac{1}{2}\left[2-(a-1)^2-(b-1)^2-(a-b)^2\right]\leq 1$$
And equality hold when 
$$(a-1)^2=0\Rightarrow a=1\Rightarrow 3^x=1=3^0$$ and $$(b-1)^2=0\Rightarrow b=1\Rightarrow 5^x=1=5^0$$ and $$(a-b)^2=0\Rightarrow a=b\Rightarrow 3^x=5^x$$
So from above we get $x=0$ for which $f(x)=3^x+5^x-4^x+15^x-9^x.$ is maximum.
And $\max{f(x)}=1$ at $x=0$ 
A: $$f(x)-1=3^x + 5^x - 9^x + 15^x - 25^x-1=-\frac{1}{2}\left( (3^x-1)^2+(5^x-3^x)^2+(1-5^x)^2\right) \leq 0$$
Second solution By C-S inequality
$$(3^x+5^x+15^x)^2 \leq (1 +5^x+3^x)( 3^x+1+5^x) $$
