Which of the following statements are correct?
(A) \(\displaystyle\frac{a}{b} + \frac{b}{a} \ge 2\) for all positive real numbers \(a\) and \(b\).
(B) \(\displaystyle x + \frac{1}{x} \ge 2\) for every non-zero real number \(x\).
(C) \(\displaystyle (a + b)\left(\frac{1}{a} + \frac{1}{b}\right) \ge 4\) for all positive real numbers \(a\) and \(b\).
(D) \(a^{2} + b^{2} \ge 2ab\) for all real numbers \(a\) and \(b\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (C) and (D) only
The square of a real number is never negative.
\((a - b)^{2} \ge 0\)
\(a^{2} - 2ab + b^{2} \ge 0\)
\(a^{2} + b^{2} \ge 2ab\)
For positive \(a\) and \(b\), dividing by the positive number \(ab\) keeps the sign:
\(\displaystyle \frac{a}{b} + \frac{b}{a} \ge 2\)
So \(a^{2} + b^{2} \ge 2ab\) holds for all real numbers, and \(\frac{a}{b} + \frac{b}{a} \ge 2\) holds when \(a\) and \(b\) are positive.
Checking the statements
(A) \(\frac{a}{b} + \frac{b}{a} \ge 2\) for all positive real numbers \(a\) and \(b\). — Correct. It is \(a^{2} + b^{2} \ge 2ab\) divided by the positive number \(ab\).
(B) \(x + \frac{1}{x} \ge 2\) for every non-zero real number \(x\). — Incorrect. For \(x = -1\), \(x + \frac{1}{x} = -2\).
(C) \((a + b)\left(\frac{1}{a} + \frac{1}{b}\right) \ge 4\) for all positive real numbers \(a\) and \(b\). — Correct. It expands to \(2 + \frac{a}{b} + \frac{b}{a}\), which is at least \(2 + 2 = 4\).
(D) \(a^{2} + b^{2} \ge 2ab\) for all real numbers \(a\) and \(b\). — Correct. It is the same as \((a - b)^{2} \ge 0\).