Q.The relation between arithmetic mean and geometric mean of two numbers always be:

The AM-GM inequality states that for two non-negative reals $a,b$: $A=\dfrac{a+b}{2} \ge G=\sqrt{ab}$.

Proof sketch: $A - G = \dfrac{a+b}{2} - \sqrt{ab} = \dfrac{a+b-2\sqrt{ab}}{2} = \dfrac{(\sqrt{a}-\sqrt{b})^2}{2} \ge 0$, since a square is always $\ge 0$. …

Geometric mean $G=\sqrt{ab}$; the expression $\dfrac{a+b}{2}$ given in the statement is actually the arithmetic mean, not the geometric mean.

By definition, for two positive numbers $a$ and $b$, the geometric mean is $G=\sqrt{ab}$.

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For two different positive numbers, the arithmetic mean always exceeds the geometric mean.

For two positive numbers $x \neq y$: $A = \dfrac{x+y}{2}$, $G = \sqrt{xy}$.

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The geometric mean (GM) of two numbers $a$ and $b$ is $\sqrt{ab}$; for 2 and 8 this gives 4.

The geometric mean of two positive numbers $a$ and $b$ is defined as $G = \sqrt{ab}$.

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The arithmetic mean gives the sum of the two numbers, and the geometric mean gives their product; solve the resulting quadratic to find the numbers.

Let the two numbers be $a$ and $b$.

Arithmetic mean: $\dfrac{a+b}{2} = 10 \implies a+b = 20$

Geometric mean: $\sqrt{ab} = 8 \implies ab = 64$

So $a$ and $b$ are roots of the quadratic $t^2 - (\text{sum})t + (\text{product}) = 0$:

$t^2 - 20t + 64 = 0$

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The geometric mean of two positive numbers $a$ and $b$ is $\sqrt{ab}$.

Step 1. If $a, G, b$ are in G.P., then $\dfrac{G}{a} = \dfrac{b}{G}$, so $G^2 = ab$.

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The AM-GM inequality states $AM \ge GM$ for any two positive numbers, always.

For two positive numbers $a,b$: $AM=\dfrac{a+b}{2}$, $GM=\sqrt{ab}$. It's a standard, provable inequality that:

$\frac{a+b}{2} \ge \sqrt{ab}$

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The expression equals the Arithmetic Mean $\frac{a+b}{2}$ only when $n=1$.

The Arithmetic Mean of $a$ and $b$ is $\frac{a+b}{2}$.

We need $\dfrac{a^n+b^n}{a^{n-1}+b^{n-1}} = \dfrac{a+b}{2}$.

Try $n = 1$: $\dfrac{a^1+b^1}{a^0+b^0} = \dfrac{a+b}{1+1} = \dfrac{a+b}{2}$, which is exactly the Arithmetic Mean.

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$f(x)\ge 2$; option (d).

$f(x)=\cos^2x+\sec^2x=\cos^2x+\dfrac{1}{\cos^2x}.$

For any positive quantity $t$, $t+\frac1t\ge 2$ (AM–GM), with equality when $t=1$. Here $t=\cos^2x$, so …

Since the two positive numbers are different, the AM-GM inequality is strict: $A > G$.

For two positive numbers $a, b$: arithmetic mean $A = \dfrac{a+b}{2}$, geometric mean $G = \sqrt{ab}$.

The AM-GM inequality states $A \ge G$ always, with equality only when $a = b$.

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By the AM–GM inequality, A > G strictly whenever a ≠ b.

For two positive numbers $a, b$: $A = \dfrac{a+b}{2}$ (arithmetic mean), $G = \sqrt{ab}$ (geometric mean).

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The statement mixes up AM and GM; the correct formula for AM is $\frac{a+b}{2}$, so the statement as given is False.

For two positive real numbers $a$ and $b$:

• Arithmetic Mean (AM) $= \dfrac{a+b}{2}$
• Geometric Mean (GM) $= \sqrt{ab}$ …