Roots of the equation $(1-4x)^4+32x^4=\frac{1}{27}​$

Question

find all real roots of the equation $(1-4x)^4+32x^4=\dfrac{1}{27}​$

i try to use binomial expansion and am-gm inequality but i don't know how to do next.

Answer 1

By curiosity, let us find the minimum of the LHS.

From the derivative, we solve

$$(1-4x)^3=8x^3$$ or $$1-4x=2x$$ so that there is a single local minimum at

$$x=\frac16$$ and $$f(x)=\frac1{81}+\frac2{81}\ !$$

Answer 2

First, we can rewrite the left hand side of your equation as: $$\text{LHS} \stackrel{def}{=}(1-4x)^4 + (2x)^4 + (2x)^4$$

Notice $(1-4x) + (2x) + (2x) = 1$ and the map $t \mapsto t^4$ is strictly convex. By Jensen's inequality, we have $$\text{LHS} \ge 3\left[\frac{(1-4x)+(2x)+(2x)}{3}\right]^4 = \frac1{27}$$ and the equality is reached when and only when $2x = 1-4x$.

This implies your original equation has one (and only one) real solution: $x = \frac16$.

Answer 3

Clearly there are no solutions if $ x > \frac{1}{4}$ or $ x < 0$. So, we may apply the power mean inequality to $ 1-4x, 2x, 2x$.

$$\sqrt[4]{\frac{( 1- 4x)^ 4 + (2x)^4 + (2x)^4 }{3}}\geq \frac{ (1-4x) + 2x + 2x } { 3} = \frac{1}{3} $$

This tells us that $( 1- 4x)^ 4 + 32x^4 \geq \frac{1}{27}$.

Equality holds iff $ 1 -4x = 2x = 2x$, or that $ x = \frac{1}{6}$.

Answer 4

If $x<0$ so $$(1-4x)^4+32x^4>1>\frac{1}{27}.$$ If $x>\frac{1}{4}$ so $$(1-4x)^4+32x^4>32\left(\frac{1}{4}\right)^4>\frac{1}{27}.$$ Id est, it's enough to solve our equation for $x\in\left[0,\frac{1}{4}\right]$.

But in this case by Holder we obtain: $$(1-4x)^4+32x^4=\frac{1}{27}(1+2)^3((1-4x)^4+2(2x)^4)\geq\frac{1}{27}(1-4x+2\cdot2x)^4=\frac{1}{27}.$$ But we have equality and the equality in Holder occurs for $(1,2)||((1-4x)^4,32x^4)$ only,

which gives $$1-4x=2x$$ or $$x=\frac{1}{6}.$$