Who needs classical inequalities?

Assuming arithmetic mean... $\frac{A+B}{2}=C$ $\implies C^2=\frac{A^2+2AB+B^2}{4}$ $\frac{A^2+B^2}{2}=\frac{A^2+2AB+B^2}{4}$ $\implies 2A^2+2B^2=A^2+2AB+B^2$ $\implies 0=-A^2+2AB-B^2$ $\implies 0=-(A^2-2AB+B^2)$ $\implies 0=(A-B)^2$

However, $A$ and $B$ are distinct. Therefore, their difference can not be 0 and the situation is impossible. $~\beta_{\lceil \mid \rceil}$

We are trying to find a solution to

$A^2+B^2=2C^2 \\ A+B=2C$

Dividing the equations by $C^2$ and $C$ respectively, we see this is equivalent to solving

$\alpha^2+\beta^2=2 \\ \alpha+\beta=2$

In the alpha-beta plane this amounts to the intersection of a circle with a line. This can have at most two solutions, and it is easy to confirm with a quick sketch that in fact the line is a tangent to the circle and there is one (repeated) solution at (1,1).

Tracing the argument backwards we see that this solution results in A = B.

Since the problem stipulated that A and B were distinct positive real numbers, we see that no such solutions exist to the problem.

Relevant wiki: Cauchy-Schwarz Inequality

The average of $A$ and $B$ is given by $C=\dfrac {A+B}2$ . The average $A^2$ and $B^2$ is $\dfrac {A^2+B^2}2$ . If $\dfrac {A^2+B^2}2 = C^2$ , then

$\begin{aligned} \frac {A^2+B^2}2 & = \left(\frac {A+B}2\right)^2 \\ \implies (A+B)^2 & = 2 (A^2+B^2) \end{aligned}$

By Cauchy-Schwarz inequality, we have $(A+B)^2 \le 2 (A^2+B^2)$ for all positive $A$ and $B$ and equality occurs when $A=B$ . Since $A$ and $B$ are distinct, equality cannot occur. No, it is not possible ,

Assume it's possible.

$\dfrac{A + B}{2} = C \implies \dfrac{A^2 + B^2}{2} = (\dfrac{A + B}{2})^2 \implies 2A^2 + 2B^2 = A^2 + 2AB + B^2 \implies (A - B)^2 = 0 \implies A = B$ .

Since $A$ and $B$ are given as distinct positive numbers $\implies$ it's not possible.

Let $x$ be a real number such that $A = C + x$ and $B = C - x$ . This ensures $\dfrac{A + B} {2} = C$ . But: $\dfrac{A^2 + B^2}{2} = \dfrac{(C + x)^2 +(C-x) ^2}{2} =\dfrac{C^2 + 2Cx + x^2 + C^2 - 2Cx + x^2}{2} = C^2 + x^2 = C^2 \Rightarrow \boxed{x = 0}$

This implies that $A = B = C$ which is contradictional, as $A$ and $B$ must be distinct positive numbers.

Note that $\frac{A+B}{2}=C$ $\frac{(A+B)^2}{4}=C^2$ $\frac{A^2+B^2+2AB}{4}=C^2$ $A^2+B^2+2AB=4C^2$ If $\frac{A^2+B^2}{2}=C^2$ , then $A^2+B^2=2C^2$ , and we have $A^2+B^2+2AB=4C^2$ $2C^2+2AB=4C^2$ $2C^2+2AB=4C^2$ $2AB=2C^2$ $AB=C^2$ We know that $AB$ is always positive, as a positive number times a positive number is positive. $C^2$ is also always positive, as the average of two positive numbers must be positive. This means that we can square root both sides. Doing so gives $\sqrt{AB}=C$ . By AM-GM, $\sqrt{AB}\le\frac{A+B}{2}=C$ . Equality is achieved when $A=B$ , but we are given that $A$ and $B$ are distinct, so this is impossible.

This is only possible when the arithmatic and geometric mean are same, which is true only for two identical numbers. But the constraint is that they are distinct, hence imposible.

We have: $\begin{cases} a + b = 2c \\ a^2 + b^2= 2c^2 \end{cases}$

$\Leftrightarrow$ $\begin{cases} a^2 + 2ab + b^2 = 4c^2 \\ a^2 + b^2= 2c^2 \end{cases}$

$\Leftrightarrow$ $\begin{cases} a + b = 2c \\ 2ab = 2c^2 \end{cases}$

$\Leftrightarrow$ $\begin{cases} \frac{a+b}{2} = c \\ \sqrt{ab} = c \end{cases}$

or $\frac{a+b}{2} = \sqrt{ab}$

According to the Cauchy theorem, this happens if and only if a = b, which contradict the statement. So the answer is No, it's not possible for the average of $a^2$ and $b^2$ to equals to $c^2$ .

Note that $\displaystyle \frac{A^2+B^2}{2} \ge \left(\frac{A+B}{2}\right)^2 =C^2$ with the equality holds if and only if $A=B$ . Since $A \neq B$ , $\displaystyle \frac{A^2+B^2}{2} > C^2$ , which means that the average of $A^2$ and $B^2$ is always greater than $C^2$ .

Quadratic function is convex, therefore, for two differences positive numbers square of the average will be always less than average of squares

For A=8,B=7andC=7.5 A+B/2=7.5 But A^2+B^2/2 is not equal to C^2 I.E,88.5 is not equal to 56.25

No bcuz I tried 2 different numbers and it didn’t work