Ta có : \(\sqrt{2015a+bc}=\sqrt{\left(a+b+c\right)a+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\)

Áp dụng BĐT Bu-nhi-a-cốp-ski, ta có : \(\left(a+b\right)\left(a+c\right)=\left(\sqrt{a}^2+\sqrt{b}^2\right)\left(\sqrt{a}^2+\sqrt{c}^2\right)\ge\left(\sqrt{ac}+\sqrt{ab}\right)^2\)

\(\Rightarrow\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{ac}+\sqrt{ab}\)

\(\Rightarrow\frac{a}{a+\sqrt{2015a+bc}}\le\frac{a}{a+\sqrt{ac}+\sqrt{ab}}=\frac{\sqrt{a}^2}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

\(\Rightarrow\Sigma\frac{a}{a+\sqrt{2015a+bc}}\le\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\)

bạn thế 2019=a+b+c de thoi ma

Ta có: \(2019a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(c+a\right)\ge\left(\sqrt{ab}+\sqrt{ac}\right)^2\)

\(\Rightarrow a+\sqrt{2019a+bc}\ge a+\sqrt{ab}+\sqrt{bc}=\sqrt{a}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

\(\Rightarrow\frac{a}{a+\sqrt{2019a+bc}}\le\frac{a}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Tương tự cộng vào suy ra điều phải chứng minh

Mình ra rồi tks mấy bn

Ta có : \(\frac{a}{a+\sqrt{2013a+bc}}=\frac{a}{a+\sqrt{a^2+ab+ac+bc}}=\frac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\)

Theo bất đẳng thức Bunhiacopxki : \(\sqrt{\left(a+b\right)\left(c+a\right)}\ge\sqrt{\left(\sqrt{ac}+\sqrt{ab}\right)^2}=\sqrt{ab}+\sqrt{ac}\)

\(\Rightarrow\frac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{a}{a+\sqrt{ab}+\sqrt{ac}}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

hay \(\frac{a}{a+\sqrt{2013a+bc}}\le\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Tương tự : \(\frac{b}{b+\sqrt{2013b+ac}}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

\(\frac{c}{c+\sqrt{2013c+ab}}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Cộng các bất đẳng thức trên theo vế được \(\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ac}}+\frac{c}{c+\sqrt{2013c+ab}}\le1\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\\a+b+c=2013\\a,b,c>0\end{cases}}\) \(\Leftrightarrow a=b=c=671\)

v

Đặt \(M=\sqrt{2015a+1}+\sqrt{2015b+1}+\sqrt{2015c+1}\)

\(\Rightarrow M^2\le\left(1+1+1\right)\left(2015a+1+2015b+1+2015c+1\right)\) (bđt Cauchy Shwarz)

\(=6048\) \(\left(a+b+c=1\right)\)

\(\Rightarrow M\le\sqrt{6048}< \sqrt{6084}=78\) (đpcm)

Để dễ nhìn, đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\)

\(VT=\frac{xy}{z^2+2xy}+\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}\)

\(2VT=\frac{2xy}{z^2+2xy}+\frac{2yz}{x^2+2yz}+\frac{2zx}{y^2+2xz}=1-\frac{z^2}{z^2+2xy}+1-\frac{x^2}{x^2+2yz}+1-\frac{y^2}{y^2+2xz}\)

\(2VT=3-\left(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\right)\)

\(2VT\le3-\frac{\left(x+y+z\right)^2}{x^2+2yz+y^2+2xz+z^2+2xy}=3-\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=2\)

\(\Rightarrow VT\le1\)

Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)