cmi-entrance 2024 QB3

cmi-entrance · India · ugmath 15 marks Proof Direct Proof of an Inequality
(a) For non-negative numbers $a, b, c$ and any positive real number $r$ prove the following inequality and state precisely when equality is achieved. $$a^r(a-b)(a-c) + b^r(b-a)(b-c) + c^r(c-a)(c-b) \geq 0$$ Hint: Assuming $a \geq b \geq c$ do algebra with just the first two terms. What about the third term? What if the assumption is not true?
(b) As a special case obtain an inequality with $a^4 + b^4 + c^4 + abc(a+b+c)$ on one side.
(c) Show that if $abc = 1$ for positive numbers $a, b, c$, then $$a^4 + b^4 + c^4 + a^3 + b^3 + c^3 + a + b + c \geq \frac{a^2+b^2}{c} + \frac{b^2+c^2}{a} + \frac{c^2+a^2}{b} + 3.$$
(a) For non-negative numbers $a, b, c$ and any positive real number $r$ prove the following inequality and state precisely when equality is achieved.
$$a^r(a-b)(a-c) + b^r(b-a)(b-c) + c^r(c-a)(c-b) \geq 0$$
Hint: Assuming $a \geq b \geq c$ do algebra with just the first two terms. What about the third term? What if the assumption is not true?

(b) As a special case obtain an inequality with $a^4 + b^4 + c^4 + abc(a+b+c)$ on one side.

(c) Show that if $abc = 1$ for positive numbers $a, b, c$, then
$$a^4 + b^4 + c^4 + a^3 + b^3 + c^3 + a + b + c \geq \frac{a^2+b^2}{c} + \frac{b^2+c^2}{a} + \frac{c^2+a^2}{b} + 3.$$