Ap dông B§T C-S ta cã:

\(\frac{a}{a+\sqrt{2016a+bc}}=\frac{a}{a+\sqrt{\left(a+b+c\right)a+bc}}=\frac{a}{a+\sqrt{\left(a+b\right)\left(c+a\right)}}\)

\(\le\frac{a}{a+\sqrt{\left(\sqrt{ab}+\sqrt{ac}\right)^2}}=\frac{a}{a+\sqrt{ab}+\sqrt{ac}}\)

\(=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\). T­uong tù ta cx cã: 

\(\frac{b}{b+\sqrt{2016b+ca}}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}};\frac{c}{c+\sqrt{2016c+ab}}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Céng theo vÕ c¸c B§T trªn ta dc:

\(VT\le\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\)

P/s:may mk bi loi Unikey r` mk dg ban chua kip chinh lai bn gang doc 

Áp dụng BĐT Cauchy-Schwarz:

$\frac{a}{a+\sqrt{2016a + bc}}=\frac{a}{a+\sqrt{(a+b+c)a + bc}} =\frac{a}{a+\sqrt{(a+b)(c+a)}} \leq \frac{a}{a+\sqrt{(\sqrt{ab}+\sqrt{ac})^{2}}}=\frac{a}{a+\sqrt{ab}+\sqrt{ac}}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$

$\Rightarrow \frac{a}{a+\sqrt{2016a + bc}} + \frac{b}{b+\sqrt{2016b + ca}} + \frac{c}{c+\sqrt{2016c + ab}}\leq \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1$

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ĐỀ thi hsg toán 9 hải phòng năm 2016-2017

Ta có:\(\sqrt{a\left(1-b\right)\left(1-c\right)}=\sqrt{a\left(1-b-c+ab\right)}\)

\(=\sqrt{a\left(a+2\sqrt{abc}+bc\right)}=\sqrt{a\left(\sqrt{a}+\sqrt{bc}\right)^2}\)

\(=\sqrt{\left(a+\sqrt{abc}\right)^2}=a+\sqrt{abc}\)

Tương tự ta CM dc:

\(\sqrt{b\left(1-c\right)\left(1-a\right)}=b+\sqrt{abc};\sqrt{c\left(1-a\right)\left(1-b\right)}=c+\sqrt{abc}\)

\(\Rightarrow P=a+\sqrt{abc}+b+\sqrt{abc}+c+\sqrt{abc}-\sqrt{abc}+2016\)

\(P=a+b+c+2\sqrt{abc}+2016\)

\(P=1+2016=2017\)