A.M. – G.M. – H.M. Inequality — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics A.M. – G.M. – H.M. Inequality MCQs with step-by-step solutions (20 questions). Part of Sequences and Series. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — A.M. – G.M. – H.M. Inequality · easy · theory
For two distinct positive numbers, the correct ordering of the means is:
A. $A.M. > G.M. > H.M.$ ✓ Correct
B. $G.M. > A.M. > H.M.$
C. $A.M. = G.M. = H.M.$
D. $H.M. > G.M. > A.M.$
Solution: For distinct positive numbers, $A > G > H$; equality holds only when the numbers are equal.
Q2 — A.M. – G.M. – H.M. Inequality · medium · numerical
For two positive numbers, if $A.M. = 9$ and $H.M. = 4$, then the G.M. is:
A. $5$
B. $\sqrt{13}$
C. $6$ ✓ Correct
D. $6.5$
Solution: Since $G^2 = A \cdot H = 36$, we get $G = 6$.
Q3 — A.M. – G.M. – H.M. Inequality · medium · numerical
If the A.M. of two numbers is $10$ and their H.M. is $8$, their G.M. is:
A. $9$
B. $\sqrt{18}$
C. $4\sqrt{5}$ ✓ Correct
D. $8\sqrt{2}$
Solution: $G = \sqrt{A \cdot H} = \sqrt{80} = 4\sqrt5$.
Q4 — A.M. – G.M. – H.M. Inequality · easy · theory
For two positive numbers, the three means $A, G, H$ always satisfy:
A. $G^2 = A \cdot H$ ✓ Correct
B. $A = G = H$
C. $H^2 = A \cdot G$
D. $A^2 = G \cdot H$
Solution: The G.M. is the geometric mean of the A.M. and H.M.: $G^2 = AH$, so $A, G, H$ are in G.P.
Q5 — A.M. – G.M. – H.M. Inequality · medium · numerical
For two positive numbers with $A.M. = 5$ and $G.M. = 4$, the H.M. is:
A. $4.5$
B. $3.2$ ✓ Correct
C. $2.5$
D. $4$
Solution: $H = \dfrac{G^2}{A} = \dfrac{16}{5} = 3.2$.
Q6 — A.M. – G.M. – H.M. Inequality · hard · numerical
The A.M., G.M. and H.M. of $3$ and $27$ are respectively:
A. $9, 15, 5.4$
B. $15, 5.4, 9$
C. $15, 9, 5.4$ ✓ Correct
D. $12, 9, 6$
Solution: $A = 15,\ G = \sqrt{81} = 9,\ H = \dfrac{2 \cdot 81}{30} = 5.4$; indeed $A > G > H$ and $G^2 = AH$.
Q7 — A.M. – G.M. – H.M. Inequality · hard · numerical
For positive $a, b$, the minimum value of $(a + b)\left(\dfrac1a + \dfrac1b\right)$ is:
A. $4$ ✓ Correct
B. $1$
C. $2$
D. $8$
Solution: Expanding gives $2 + \tfrac ab + \tfrac ba \ge 2 + 2 = 4$ (A.M.–G.M.), attained at $a = b$.
Q8 — A.M. – G.M. – H.M. Inequality · medium · numerical
If $a, b, c > 0$ with $abc = 1$, the minimum value of $a + b + c$ is:
A. $9$
B. $3$ ✓ Correct
C. $\tfrac13$
D. $1$
Solution: By A.M.–G.M., $a + b + c \ge 3(abc)^{1/3} = 3$, attained at $a = b = c = 1$.
Q9 — A.M. – G.M. – H.M. Inequality · hard · numerical
For $x > 0$, the minimum value of $\dfrac{1 + x + x^2}{x}$ is:
A. $2$
B. $3$ ✓ Correct
C. $4$
D. $1$
Solution: It equals $\tfrac1x + 1 + x \ge 1 + 2 = 3$, using $x + \tfrac1x \ge 2$.
Q10 — A.M. – G.M. – H.M. Inequality · medium · numerical
The minimum value of $\tan^2\theta + \cot^2\theta$ is:
A. $0$
B. $1$
C. $\tfrac12$
D. $2$ ✓ Correct
Solution: By A.M.–G.M., $\tan^2\theta + \cot^2\theta \ge 2\sqrt{\tan^2\theta \cot^2\theta} = 2$.
Q11 — A.M. – G.M. – H.M. Inequality · hard · numerical
If $a + b = 1$ with $a, b > 0$, the minimum value of $\dfrac1a + \dfrac1b$ is:
A. $1$
B. $8$
C. $2$
D. $4$ ✓ Correct
Solution: $\tfrac1a + \tfrac1b = \dfrac{a+b}{ab} = \dfrac{1}{ab}$; since $ab \le \tfrac14$, the minimum is $4$ at $a = b = \tfrac12$.
Q12 — A.M. – G.M. – H.M. Inequality · hard · numerical
For $x > 0$, the minimum value of $\dfrac{(x+2)(x+8)}{x}$ is:
A. $16$
B. $20$
C. $18$ ✓ Correct
D. $10$
Solution: Expanding: $x + 10 + \tfrac{16}{x} \ge 10 + 2\sqrt{16} = 18$, attained at $x = 4$.
Q13 — A.M. – G.M. – H.M. Inequality · medium · numerical
The maximum value of $\sin\theta\cos\theta$ is:
A. $\tfrac12$ ✓ Correct
B. $\tfrac14$
C. $2$
D. $1$
Solution: $\sin\theta\cos\theta = \tfrac12\sin 2\theta$, whose maximum value is $\tfrac12$.
Q14 — A.M. – G.M. – H.M. Inequality · medium · theory
The equality $A.M. = G.M. = H.M.$ for a set of positive numbers holds if and only if:
A. all the numbers are equal ✓ Correct
B. the numbers are consecutive
C. exactly two numbers are equal
D. the numbers are in G.P.
Solution: The three means coincide precisely when every number in the set is the same.
Q15 — A.M. – G.M. – H.M. Inequality · medium · numerical
If $a + b + c = 15$ with $a, b, c > 0$, the maximum value of $abc$ is:
A. $75$
B. $125$ ✓ Correct
C. $100$
D. $225$
Solution: By A.M.–G.M., $abc \le \left(\tfrac{15}{3}\right)^3 = 125$, attained at $a = b = c = 5$.
Q16 — A.M. – G.M. – H.M. Inequality · hard · numerical
For two positive numbers with A.M. $= 34$ and G.M. $= 16$, the H.M. is:
A. $8$
B. $\dfrac{128}{17}$ ✓ Correct
C. $16$
D. $\dfrac{17}{128}$
Solution: $H = \dfrac{G^2}{A} = \dfrac{256}{34} = \dfrac{128}{17}$.
Q17 — A.M. – G.M. – H.M. Inequality · medium · numerical
The minimum value of $x^2 + y^2$ subject to $x + y = 8$ (real $x, y$) is:
A. $64$
B. $32$ ✓ Correct
C. $16$
D. $8$
Solution: By the power-mean (or A.M.–Q.M.) inequality the minimum is at $x = y = 4$: $16 + 16 = 32$.
Q18 — A.M. – G.M. – H.M. Inequality · hard · theory
For distinct positive numbers, the G.M. lies:
A. equal to the A.M.
B. strictly between the H.M. and the A.M. ✓ Correct
C. above the A.M.
D. below the H.M.
Solution: Since $H < G < A$ for distinct positives, the G.M. is squeezed between the H.M. and A.M.
Q19 — A.M. – G.M. – H.M. Inequality · hard · numerical
For positive reals $a, b, c$, the minimum value of $\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}$ is:
A. $1$
B. $3$ ✓ Correct
C. $6$
D. $9$
Solution: By A.M.–G.M., $\dfrac ab + \dfrac bc + \dfrac ca \ge 3\sqrt[3]{1} = 3$, attained at $a = b = c$.
Q20 — A.M. – G.M. – H.M. Inequality · medium · numerical
For $x > 0$, the minimum value of $16x + \dfrac{1}{x}$ is:
A. $2$
B. $4$
C. $8$ ✓ Correct
D. $16$
Solution: $16x + \tfrac1x \ge 2\sqrt{16} = 8$, attained at $x = \tfrac14$.