Točno
13. kolovoza 2016. 18:54 (7 godine, 11 mjeseci)
Determine the lowest possible value of the expression
![\frac{1}{a + x} + \frac{1}{a + y} + \frac{1}{b + x} + \frac{1}{b + y} \text{,}](/media/m/9/9/d/99d35622e30515e95f323fb393fd8448.png)
where
![a](/media/m/6/d/2/6d2832265560bb67cf117009608524f6.png)
,
![b](/media/m/e/e/c/eec0d7323095a1f2101fc1a74d069df6.png)
,
![x](/media/m/f/1/8/f185adeed9bd346bc960bca0147d7aae.png)
, and
![y](/media/m/c/c/0/cc082a07a517ebbe9b72fd580832a939.png)
are positive real numbers satisfying the inequalities
%V0
Determine the lowest possible value of the expression $$
\frac{1}{a + x} + \frac{1}{a + y} + \frac{1}{b + x} + \frac{1}{b + y} \text{,}
$$ where $a$, $b$, $x$, and $y$ are positive real numbers satisfying the inequalities $$
\frac{1}{a + x} \geq \frac12, \quad
\frac{1}{a + y} \geq \frac12, \quad
\frac{1}{b + x} \geq \frac12, \quad \text{and}
\ \frac{1}{b + y} \geq 1 \text{.}
$$
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![\frac{1}{a+x} + \frac{1}{a+y} + \frac{1}{b+x} + \frac{1}{b+y} \geq \frac{1}{a+x} + \frac{1}{b+y} + \frac{4}{(a+x)+(b+y)} \geq \frac{1}{2} + 1 + \frac{4}{3} = \frac{17}{6}](/media/m/4/2/0/420359f3d10e2e4a2ca18240c08537e8.png)
gdje smo koristili AH nejednakost na članove
![\frac{1}{a+y}](/media/m/7/7/2/772d212b9eb07c03fb93b043a65ef2fc.png)
i
![\frac{1}{b+x}](/media/m/7/2/c/72cabce13d0fa4aee322dd6c5f0de0bb.png)
te iz uvjeta zadatka nejednakosti
![a+x \leq 2](/media/m/5/3/0/53097f202fa6d1946b2788ec99159692.png)
i
![b+y \leq 1](/media/m/7/6/8/7686fd8b56ffe0b7925da505473f46f4.png)
. Jednakost se postiže za npr.
![(a,b,x,y)=(1,\frac{1}{2},1,\frac{1}{2})](/media/m/1/6/2/162b230b524cd7b153094fbeed9048e8.png)
.
%V0
$$\frac{1}{a+x} + \frac{1}{a+y} + \frac{1}{b+x} + \frac{1}{b+y} \geq \frac{1}{a+x} + \frac{1}{b+y} + \frac{4}{(a+x)+(b+y)} \geq \frac{1}{2} + 1 + \frac{4}{3} = \frac{17}{6}$$
gdje smo koristili AH nejednakost na članove $\frac{1}{a+y}$ i $\frac{1}{b+x}$ te iz uvjeta zadatka nejednakosti $a+x \leq 2$ i $b+y \leq 1$. Jednakost se postiže za npr. $(a,b,x,y)=(1,\frac{1}{2},1,\frac{1}{2})$.
15. rujna 2016. 21:49 | grga | Točno |