(x+y)^3+4xy>=2 tính S=x+y
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B2
50
DL
0
NL
1
28 tháng 11 2018
\(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=3-\sqrt{3}+\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1}\)
\(=3-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=3-\sqrt{3}+\sqrt{3}-1\)
\(=2\)
KT
2
TN
28 tháng 11 2018
\(P=\frac{\sqrt{ab}}{c+3\sqrt{ab}}+\frac{\sqrt{bc}}{a+3\sqrt{bc}}+\frac{\sqrt{ca}}{b+3\sqrt{ca}}\)
\(=1-\frac{1}{3}\left(\frac{c^2}{c^2+3c\sqrt{ab}}+\frac{a^2}{a^2+3a\sqrt{bc}}+\frac{b^2}{b^2+3b\sqrt{ca}}\right)\)
\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}\right)\)
\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\right)\)\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\frac{\left(a+b+c\right)^2}{3}}\right)\)
\(\le1-\frac{1}{3}\cdot\frac{\left(a+b+c\right)^2}{\frac{4\left(a+b+c\right)^2}{3}}=1-\frac{1}{4}=\frac{3}{4}\)
HA
14 tháng 3 2020
\(P=\frac{\sqrt{ab}}{c+3\sqrt{ab}}+\frac{\sqrt{bc}}{a+3\sqrt{bc}}+\frac{\sqrt{ca}}{b+3\sqrt{ca}}\)
\(=1-\frac{1}{3}\left(\frac{c^2}{c^2+3c\sqrt{ab}}+\frac{a^2}{a^2+3a\sqrt{bc}}+\frac{b^2}{b^2+3b\sqrt{ca}}\right)\)
\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}\right)\)
\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\right)\)
\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\frac{\left(a+b+c\right)^2}{3}}\right)\)
\(\le1-\frac{1}{3}\cdot\frac{\left(a+b+c\right)^2}{\frac{4\left(a+b+c\right)^2}{3}}\)
\(=1-\frac{1}{4}\)
\(=\frac{3}{4}\)