Question

Which of the following inequalities is true for any positive real numbers a, b and c?
I. \( ab(a+b)+bc(b+c)+ca(c+a) \le 6abc \)
II. \( \frac{a^{2}+b^{2}+c^{2}}{abc} \le \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \)

• A. Only I
• B. Only II
• C. Neither I nor II
• D. Both I and II

Hint:

Recall standard inequalities: \( a^2+b^2+c^2 \ge ab+bc+ca \) and the AM-GM sum \( \sum_{cyc} a^2b \ge 6abc \). If the question flips the sign (\(\le\) instead of \(\ge\)), the statement is likely false.

Answer 1

The Correct Option is C

Step 1: Analyze Inequality I:
\[ ab(a+b)+bc(b+c)+ca(c+a) \le 6abc \] By AM-GM inequality, for positive numbers, \( a+b \ge 2\sqrt{ab} \), etc. However, a standard result is \( (a+b)(b+c)(c+a) \ge 8abc \). Consider the expanded form \( a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 \). Using AM-GM on these 6 terms: \[ \frac{\text{Sum}}{6} \ge \sqrt[6]{a^6b^6c^6} = abc \implies \text{Sum} \ge 6abc \] The inequality given is \( \le 6abc \). Since the sum is actually \( \ge 6abc \), statement I is generally false (it only holds if \( a=b=c \), otherwise the LHS is strictly greater). Step 2: Analyze Inequality II:
\[ \frac{a^{2}+b^{2}+c^{2}}{abc} \le \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \] Simplify LHS: \( \frac{a}{bc} + \frac{b}{ac} + \frac{c}{ab} \). Simplify RHS: \( \frac{bc+ac+ab}{abc} = \frac{ab+bc+ca}{abc} \). So the inequality becomes: \( a^2+b^2+c^2 \le ab+bc+ca \). We know that for any real numbers, \( a^2+b^2+c^2 \ge ab+bc+ca \) (derived from \( (a-b)^2+(b-c)^2+(c-a)^2 \ge 0 \)). The statement claims \( \le \). This is false unless \( a=b=c \). Step 3: Conclusion:
Since both inequalities claim the reverse of the standard established inequalities, neither is true for "any" positive real numbers (specifically distinct ones).