Relation Between AM, GM and HM Complete JEE Notes with Proofs

Step 1: Count the terms. The AP is $59,A_1,\dots,A_{39},159$ — that is $41$ terms, so $40$ gaps.

Step 2: Common difference.

$d=\frac{159-59}{40}=\frac{100}{40}=2.5.$

Step 3: Write the four means. Using $A_r=59+rd$:

$A_{25}=59+62.5=121.5,\qquad A_{28}=59+70=129,$ $A_{31}=59+77.5=136.5,\qquad A_{36}=59+90=149.$

Step 4: Average them.

$\frac{121.5+129+136.5+149}{4}=\frac{536}{4}=\boxed{134}$

Faster route. The four subscripts $25,28,31,36$ average to $30$, and $A_r$ is linear in $r$, so the mean of the four values is $A_{30}=59+30(2.5)=134$ directly.

Step 1: Use the AP to name everything by one letter. With $4$ as the second term and common difference $d$:

$a=4-d,\qquad \alpha=4+d,\qquad b=4+2d.$

Step 2: Translate the mean condition.

$\frac12\left(\frac{1}{4-d}+\frac{1}{4+2d}\right)=\frac{5}{16}\ \Longrightarrow\ \frac{(4+2d)+(4-d)}{(4-d)(4+2d)}=\frac58.$

Step 3: Clear and solve. With $(4-d)(4+2d)=16+4d-2d^2$,

$8(8+d)=5\left(16+4d-2d^2\right)\ \Longrightarrow\ 10d^2-12d-16=0\ \Longrightarrow\ 5d^2-6d-8=0,$

so $(5d+4)(d-2)=0$ and $d=-\dfrac45$ or $d=2$.

Step 4: Pick the branch. $d=2$ gives $a=2$, $\alpha=6$, $b=8$. (The other branch $d=-\tfrac45$ gives $a=4.8$, and both branches land the roots in the same two intervals, so the answer is unaffected.)

Step 5: Build and factorise the quadratic.

$6x^2-2x+2(6-16)=6x^2-2x-20=0\ \Longrightarrow\ 3x^2-x-10=0\ \Longrightarrow\ (3x+5)(x-2)=0.$

Step 6: Read off. $x=2$ and $x=-\dfrac53\approx-1.67$.

Answer: one root in $(1,4)$ and the other in $(-2,0)$.

Step 1: Name the numbers. Let $x_1,x_2,\dots,x_n>0$ with $x_1x_2\cdots x_n=1$.

Step 2: Apply AM–GM to all $n$ of them.

$\frac{x_1+x_2+\dots+x_n}{n}\ \ge\ \left(x_1x_2\cdots x_n\right)^{1/n}=1^{1/n}=1.$

Step 3: Multiply out.

$x_1+x_2+\dots+x_n\ \ge\ n.$

Step 4: Equality. All $x_i$ equal, i.e. every $x_i=1$, which does satisfy the product condition — so the bound is attained.

Answer: the sum is never less than $n$.

Step 1: Make the product match the sum. The sum contains $2a_n$, not $a_n$, so treat the $n$ quantities as

$a_1,\ a_2,\ \dots,\ a_{n-1},\ 2a_n,$

whose product is $2\left(a_1a_2\cdots a_n\right)=2c$.

Step 2: Apply AM–GM to those $n$ quantities.

$\frac{a_1+a_2+\dots+a_{n-1}+2a_n}{n}\ \ge\ \left(2c\right)^{1/n}.$

Step 3: Multiply out.

$a_1+a_2+\dots+a_{n-1}+2a_n\ \ge\ n\,(2c)^{1/n}.$

Step 4: Equality. All $n$ quantities equal, i.e. $a_1=\dots=a_{n-1}=2a_n$, which is achievable.

Answer: $\boxed{n\,(2c)^{1/n}}$

Step 1: Split the coefficient. Write $3a^{-3}$ as three separate terms. The expression becomes a sum of eight quantities:

$a^{-5},\ a^{-4},\ a^{-3},\ a^{-3},\ a^{-3},\ 1,\ a^{8},\ a^{10}.$

Step 2: Check their product. Add the exponents:

$-5-4-3-3-3+0+8+10=0,$

so the product is $a^0=1$. This is why the question was built with these particular powers.

Step 3: Apply AM–GM to the eight quantities.

$\frac{a^{-5}+a^{-4}+3a^{-3}+1+a^{8}+a^{10}}{8}\ \ge\ \left(1\right)^{1/8}=1.$

Step 4: Multiply out.

$a^{-5}+a^{-4}+3a^{-3}+1+a^{8}+a^{10}\ \ge\ \boxed{8}$

Step 5: Equality. All eight equal forces $a=1$, and substituting $a=1$ gives $1+1+3+1+1+1=8$. The bound is attained, so $8$ really is the minimum.

Step 1: The middle term of an AP is the average of its neighbours.

$f(a)=\frac12\Big[\left(2^{1-a}+2^{1+a}\right)+\left(3^{a}+3^{-a}\right)\Big].$

Step 2: Minimise each bracket by AM–GM. Each is a sum of two positives with constant product:

$2^{1-a}+2^{1+a}\ \ge\ 2\sqrt{2^{1-a}\cdot 2^{1+a}}=2\sqrt{2^2}=4,$ $3^{a}+3^{-a}\ \ge\ 2\sqrt{3^{a}\cdot 3^{-a}}=2.$

Both equalities need $a=0$, the same value of $a$ — so both minima are reached together.

Step 3: Minimum of $f$.

$\alpha=f(0)=\frac12(4+2)=3.$

Step 4: Set up the integral. The limits are $\log_e 2$ and $\log_e 3$. Substitute $t=e^{2x}$, so $dx=\dfrac{dt}{2t}$ and $e^{2x}-e^{-2x}=\dfrac{t^2-1}{t}$; the limits become $t=4$ and $t=9$:

$I=\int_4^9\frac{t}{t^2-1}\cdot\frac{dt}{2t}=\int_4^9\frac{dt}{2\left(t^2-1\right)}.$

Step 5: Partial fractions. Using $\dfrac{1}{t^2-1}=\dfrac12\left(\dfrac{1}{t-1}-\dfrac{1}{t+1}\right)$,

$I=\frac14\left[\log_e\frac{t-1}{t+1}\right]_4^9=\frac14\left(\log_e\frac{8}{10}-\log_e\frac{3}{5}\right).$

Step 6: Simplify.

$I=\frac14\log_e\left(\frac{8}{10}\cdot\frac{5}{3}\right)=\frac14\log_e\left(\frac43\right)$

Answer: $\boxed{\dfrac14\log_e\dfrac43}$

Step 1: Group into two blocks. Treat $(a+b)$ and $(c+d)$ as two positive numbers. Their sum is $2$.

Step 2: Apply AM–GM to the two blocks.

$\frac{(a+b)+(c+d)}{2}\ \ge\ \sqrt{(a+b)(c+d)}\ \Longrightarrow\ \frac22\ \ge\ \sqrt m\ \Longrightarrow\ m\le 1.$

Step 3: Get the lower bound. All four numbers are positive, so $a+b>0$ and $c+d>0$, giving $m>0$. It cannot reach $0$.

Step 4: Combine.

$\boxed{0<m\le 1}$

Step 5: Equality. $m=1$ needs $a+b=c+d=1$, which is easily arranged, so the upper bound is attained.

Step 1: Check both terms are positive. Wherever the expression is defined we need $x^2+x>0$, so $\sqrt{x^2+x}>0$. And $\alpha\in\left(0,\dfrac\pi2\right)$ makes $\tan\alpha>0$, so $\tan^2\alpha>0$.

Step 2: Name the pieces. Put $u=\sqrt{x^2+x}$ and $v=\dfrac{\tan^2\alpha}{\sqrt{x^2+x}}$. Their product is constant:

$uv=\tan^2\alpha.$

Step 3: Apply AM–GM to $u$ and $v$.

$\frac{u+v}{2}\ \ge\ \sqrt{uv}=\sqrt{\tan^2\alpha}=\tan\alpha$

(the positive square root, since $\tan\alpha>0$ on this interval).

Step 4: Multiply out.

$\sqrt{x^2+x}+\frac{\tan^2\alpha}{\sqrt{x^2+x}}\ \ge\ \boxed{2\tan\alpha}$

Step 5: Equality. $u=v$ means $x^2+x=\tan^2\alpha$, which has a solution for every $\alpha$ in the given range, so the bound is sharp. $\blacksquare$

Step 1: Expand the product.

$(1+a)(1+b)(1+c)=1+a+b+c+ab+bc+ca+abc.$

Step 2: Count the terms. The right-hand side is a sum of seven positive quantities after the leading $1$:

$a,\ b,\ c,\ ab,\ bc,\ ca,\ abc.$

Apply AM–GM to those seven, and their product is

$a\cdot b\cdot c\cdot ab\cdot bc\cdot ca\cdot abc=a^{4}b^{4}c^{4}.$

Step 3: Write the inequality.

$\frac{a+b+c+ab+bc+ca+abc}{7}\ \ge\ \left(a^{4}b^{4}c^{4}\right)^{1/7}.$

Step 4: Add the leading $1$. Since $1>0$,

$(1+a)(1+b)(1+c)=1+\big(a+b+c+ab+bc+ca+abc\big)>7\left(a^{4}b^{4}c^{4}\right)^{1/7}.$

The inequality is now strict because of that extra $1$, regardless of whether the seven terms are equal.

Step 5: Raise to the seventh power. Both sides are positive, and $t\mapsto t^7$ is increasing, so

$\left\{(1+a)(1+b)(1+c)\right\}^{7}>7^{7}a^{4}b^{4}c^{4}. \qquad \blacksquare$

Step 1: Set up the GP. Let the numbers be $a$ and $b$. Since $a,y,z,b$ are in GP with ratio $r$:

$y=ar,\qquad z=ar^2,\qquad b=ar^3.$

Step 2: Express $a$ and $b$ through $y$ and $z$. From $\dfrac ay=\dfrac yz=\dfrac zb$ we get

$a=\frac{y^2}{z},\qquad b=\frac{z^2}{y}.$

Step 3: Use the AM condition.

$x=\frac{a+b}{2}=\frac12\left(\frac{y^2}{z}+\frac{z^2}{y}\right)=\frac{y^3+z^3}{2yz}.$

Step 4: Rearrange.

$2xyz=y^3+z^3\ \Longrightarrow\ \frac{y^3+z^3}{xyz}=\boxed{2}$

Step 1: The two easy products. For the AP, the two means are symmetric about the centre, so

$A_1+A_2=a+b.$

For the GP, the two means multiply to give

$G_1G_2=ab.$

Step 2: The harmonic means. Reciprocals of $a,H_1,H_2,b$ form an AP with common difference $\dfrac{1}{3}\left(\dfrac1b-\dfrac1a\right)$, which after simplifying gives

$H_1=\frac{3ab}{2b+a},\qquad H_2=\frac{3ab}{b+2a}.$

Step 3: Use the AP of reciprocals. Since $\dfrac1{H_1}$ and $\dfrac1{H_2}$ are symmetric about the centre of that AP,

$\frac{1}{H_1}+\frac{1}{H_2}=\frac1a+\frac1b\ \Longrightarrow\ \frac{H_1+H_2}{H_1H_2}=\frac{a+b}{ab}=\frac{A_1+A_2}{G_1G_2}.$

Cross-multiplying gives the first equality:

$\frac{G_1G_2}{H_1H_2}=\frac{A_1+A_2}{H_1+H_2}.$

Step 4: Compute the common value.

$\frac{G_1G_2}{H_1H_2}=\frac{ab}{\dfrac{3ab}{2b+a}\cdot\dfrac{3ab}{b+2a}}=\frac{ab(2b+a)(b+2a)}{9a^2b^2}=\boxed{\frac{(2a+b)(a+2b)}{9ab}}\qquad\blacksquare$

Step 1: Name the numbers with one unknown. Let them be $ka$ and $a$ with $a>0$, so the required ratio is $k:1$.

Step 2: Write both means.

$G=\sqrt{ka\cdot a}=a\sqrt k,\qquad H=\frac{2(ka)(a)}{ka+a}=\frac{2ka}{k+1}.$

Step 3: Impose the ratio.

$\frac HG=\frac{2ka}{(k+1)\,a\sqrt k}=\frac{2\sqrt k}{k+1}=\frac45.$

Step 4: Solve for $\sqrt k$.

$10\sqrt k=4k+4\ \Longrightarrow\ 2k-5\sqrt k+2=0\ \Longrightarrow\ \sqrt k=\frac{5\pm 3}{4}=2\ \text{or}\ \frac12,$

so $k=4$ or $k=\dfrac14$.

Step 5: Read the ratio. Both give the same pair of numbers in the two possible orders.

Answer: the numbers are in the ratio $\boxed{4:1}$

Step 1: AM–GM on the three numbers.

$\frac{a+b+c}{3}\ \ge\ (abc)^{1/3}.$

Step 2: AM–GM on the three reciprocals.

$\frac{\dfrac1a+\dfrac1b+\dfrac1c}{3}\ \ge\ \left(\frac{1}{abc}\right)^{1/3}.$

Step 3: Multiply the two. Both sides of both inequalities are positive, so multiplying is legal:

$\frac{(a+b+c)}{3}\cdot\frac{\left(\dfrac1a+\dfrac1b+\dfrac1c\right)}{3}\ \ge\ (abc)^{1/3}\cdot\left(\frac{1}{abc}\right)^{1/3}=1.$

Step 4: Clear the nines.

$(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\ \ge\ \boxed{9}$

Step 5: Equality. Both AM–GM steps need all three equal, so equality holds exactly when $a=b=c$. $\blacksquare$

Check. At $a=b=c=1$ the left side is $3\times3=9$. At $a=1,b=2,c=3$ it is $6\times\dfrac{11}{6}=11>9$.