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Ruji2018252   4
N Yesterday at 9:07 AM by Victoria_Discalceata1
$a,b,c\in [0;2]$ and $a+b+c=3$ find min and max:
$$P=\dfrac{2-a^2}{a+1}+\dfrac{2-b^2}{b+1}+\dfrac{2-c^2}{c+1}$$
4 replies
Ruji2018252
Yesterday at 6:56 AM
Victoria_Discalceata1
Yesterday at 9:07 AM
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Ruji2018252
168 posts
#1
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$a,b,c\in [0;2]$ and $a+b+c=3$ find min and max:
$$P=\dfrac{2-a^2}{a+1}+\dfrac{2-b^2}{b+1}+\dfrac{2-c^2}{c+1}$$
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lbh_qys
41 posts
#2
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Define \( f(x) = \frac{2-x^2}{x + 1} \), then \( f''(x) = \frac{2}{(x+1)^3}> 0 \), then by

\[
(2, 1, 0) \succ (a, b, c) \succ (1, 1, 1)
\]
and the Karamata inequality, we know that

\[
P_{\text{max}} = f(2) + f(1) + f(0) = \frac{11}{6}
\]
and

\[
P_{\text{min}} = f(1) + f(1) + f(1) = \frac{3}{2}.
\]
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sqing
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#3
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Let $ a,b,c\in [0;2] $ and $ a+b+c=3 .$ Prove that
$$\dfrac{3}{2}\leq \dfrac{2-a^2}{a+1}+\dfrac{2-b^2}{b+1}+\dfrac{2-c^2}{c+1}\leq \dfrac{11}{6}$$Good.
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Ruji2018252
168 posts
#4
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sqing wrote:
Let $ a,b,c\in [0;2] $ and $ a+b+c=3 .$ Prove that
$$\dfrac{3}{2}\leq \dfrac{2-a^2}{a+1}+\dfrac{2-b^2}{b+1}+\dfrac{2-c^2}{c+1}\leq \dfrac{11}{6}$$Good.

How prove ??
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Victoria_Discalceata1
711 posts
#5
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A solution with tangent and secant lines.

First, since $\frac{2-a^2}{a+1}=\frac{1+(1-a)(1+a)}{a+1}=\frac{1}{a+1}+1-a$, etc. hence $P=\sum\frac{1}{1+a}$.

For the lower bound, we have $$\sum\frac{1}{a+1}\ge \sum\left(-\frac{1}{4}a+\frac{3}{4}\right)=\frac{9}{4}-\frac{3}{4}=\frac{3}{2}$$
For the upper bound assume WLOG $a\ge b\ge c$. Clearly, $a\ge 1$ and $1\ge c$.
We have two cases
$\bullet$ If $b\ge 1$ then $$\sum\frac{1}{a+1}\le\left(-\frac{1}{6}a+\frac{2}{3}\right)+\left(-\frac{1}{6}b+\frac{2}{3}\right)+\frac{1}{c+1}=-\frac{1}{6}(3-c)+\frac{4}{3}+\frac{1}{c+1}=\frac{11}{6}-\frac{c(5-c)}{6(c+1)}\le\frac{11}{6}$$$\bullet$ If $1\ge b$ then $$\sum\frac{1}{a+1}\le\frac{1}{a+1}+\left(-\frac{1}{2}b+1\right)+\left(-\frac{1}{2}c+1\right)=\frac{1}{a+1}-\frac{1}{2}(3-a)+2=\frac{11}{6}-\frac{(2-a)(3a+1)}{6(a+1)}\le\frac{11}{6}$$
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