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Question Number 202348 by York12 last updated on 24/Dec/23

Let a, b, c \in R^{+}, a + b + c = 3 prove the following inequality

\frac{{(2 a - 3)}^{2}}{b} + \frac{{(2 b - 3)}^{2}}{c} + \frac{{(2 c - 3)}^{2}}{a} ⩾ \frac{a^{2} + b^{2}}{a + b} + \frac{b^{2} + c^{2}}{b + c} + \frac{c^{2} + a^{2}}{c + a}

Answered by mrdiane last updated on 25/Dec/23

o n a 0 {(a + b + c)}^{2} = a^{2} + b^{2} + 2 (a b + a c + b c c o m m e a, b e t > 0 d o n c

{(a + b + c)}^{2} ⩾ a^{2} + b^{2} o r b ⩽ a + b

d o n c o n a {\begin{cases} {(a + b + c)}^{2} ⩾ a^{2} + b^{2} (1) \\ \frac{1}{b} ⩾ \frac{1}{a + b} (2) \end{cases}

(1) \times (2) d o n n e \frac{{(a + b + c)}^{2}}{b} ⩾ \frac{a^{2} + b^{2}}{a + b} \Leftrightarrow \frac{{(a + a - 3)}^{2}}{b} ⩾ \frac{a^{2} + b^{2}}{a + b}

p a e a n n a l o g i e o n a {\begin{cases} \frac{{(2 a - 3)}^{2}}{b} ⩾ \frac{a^{2} + b^{2}}{a + b} (3) \\ \frac{{(2 b - 3)}^{2}}{c} ⩾ \frac{b^{2} + c^{2}}{b + c} (4) \\ \frac{{(2 c - 3)}^{2}}{a} ⩾ \frac{c^{2} + a^{2}}{a + c} (5) \end{cases}

(3) + (4) + (5) d o n n e \frac{{(2 a - 3)}^{2}}{b} + \frac{(2 b - 3)}{c} + \frac{(2 c - 3)}{a} ⩾ \frac{a^{2} + b^{2}}{a + b} + \frac{b^{2} + c^{2}}{b + c} + \frac{c^{2} + a^{2}}{c + a}

Commented by York12 last updated on 25/Dec/23

Merci monsieur, mais je ne comprends pas la première ligne

Commented by York12 last updated on 25/Dec/23

{(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + a c)

\neq a^{2} + b^{2} + 2 (a b + b c + a c)

b ⩾ b ⇎ b ⩾ a + b \to (because eauality holds when a = 0 b u t a > 0 \Rightarrow The equality does not hold)

{(a + b + c)}^{2} ⩾ a^{2} + b^{2} \to (I do not see the equality case)

\frac{{(a + b + c)}^{2}}{b} ⩾ \frac{a^{2} + b^{2}}{a + b} \Leftrightarrow \frac{{(a + a - 3)}^{2}}{b} ⩾ \frac{a^{2} + b^{2}}{a + b} \to (Why)

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