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Ta có \(\sqrt{bc\left(1+a^2\right)}=\sqrt{bc+a^2bc}=\sqrt{bc+a\left(a+b+c\right)}\)
\(=\sqrt{\left(a+b\right)\left(a+c\right)}\)
Đặt BT đề cho là P
\(\Leftrightarrow P=\sum\dfrac{a}{\sqrt{bc\left(1+a^2\right)}}=\sum\sqrt{\dfrac{a}{a+b}\cdot\dfrac{a}{a+c}}\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{b+a}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\cdot3=\dfrac{3}{2}\)
Dấu \("="\Leftrightarrow a=b=c=\sqrt{3}\)
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)
\(P=\sqrt{\dfrac{yz}{x^2+1}}+\sqrt{\dfrac{zx}{y^2+1}}+\sqrt{\dfrac{xy}{z^2+1}}\)
\(P=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}+\sqrt{\dfrac{zx}{y^2+xy+yz+zx}}+\sqrt{\dfrac{xy}{z^2+xy+yz+zx}}\)
\(P=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\dfrac{zx}{\left(y+z\right)\left(x+y\right)}}+\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\)
\(P\le\dfrac{1}{2}\left(\dfrac{y}{x+y}+\dfrac{z}{x+z}\right)+\dfrac{1}{2}\left(\dfrac{z}{y+z}+\dfrac{x}{x+y}\right)+\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)=\dfrac{3}{2}\)
\(P_{max}=\dfrac{3}{2}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\) hay \(a=b=c=\sqrt{3}\)
Đề sai rồi: a,b,c > 0 thì làm sao mà có: ab + bc + ca = 0 được.
\(\sqrt{\frac{\left(a+bc\right)\left(b+ac\right)}{c+ab}}=\sqrt{\frac{\left(a^2+ab+ac+bc\right)\left(b^2+bc+ba+ac\right)}{c^2+ca+cb+ab}}=\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(b+a\right)\left(b+c\right)}{\left(c+a\right)\left(c+b\right)}}=a+b\left(a,b,c>0;a+b+c=1\right)\)
Bạn làm tương tự nha
\(\Rightarrow P=a+b+c+a+b+c=2\left(a+b+c\right)=2\)
Áp dụng BĐT Cauchy
\(\Rightarrow\left(a+b+c\right)\left(ab+bc+ac\right)\ge9abc\)
\(\Rightarrow\sqrt{\dfrac{\left(a+b+c\right)\left(ab+bc+ac\right)}{abc}}\ge3\)
\(\Rightarrow P\ge3+\dfrac{4bc}{\left(b+c\right)^2}\)
Ta cần tìm Min của \(3+\dfrac{4bc}{\left(b+c\right)^2}\)
Không mất tính tổng quát giả sử \(b\ge c\)
\(\Rightarrow b+c\le2b\)\(\Leftrightarrow\left(b+c\right)^2\le4b^2\Leftrightarrow\dfrac{4bc}{\left(b+c\right)^2}\ge\dfrac{c}{b}\)
\(b\ge c\Rightarrow\dfrac{c}{b}\ge1\)
Vậy \(3+\dfrac{4bc}{\left(b+c\right)^2}\ge4\)
Dấu đẳng thức xảy ra khi a = b = c
Áp dụng BĐT bunyakovsky và AM -GM ta có:
\(\sqrt{\dfrac{\left[a+\left(b+c\right)\right]\left[bc+a\left(b+c\right)\right]}{abc}}\ge\sqrt{\dfrac{a\left(\sqrt{bc}+b+c\right)^2}{abc}}=\dfrac{\sqrt{bc}+b+c}{\sqrt{bc}}=1+\dfrac{b+c}{\sqrt{bc}}\)
\(LHS\ge1+\dfrac{b+c}{2\sqrt{bc}}+\dfrac{b+c}{2\sqrt{bc}}+\dfrac{4bc}{\left(b+c\right)^2}\ge1+3\sqrt[3]{\dfrac{4bc\left(b+c\right)^2}{4bc\left(b+c\right)^2}}=4\)
Dấu = xảy ra khi a=b=c
\(a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(a+c\right)\)
Tương tự: \(b+ca=\left(a+b\right)\left(b+c\right)\) ; \(c+ab=\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow P=a+b+b+c+c+a=2\left(a+b+c\right)=2\)
điện thoại cùi nên chụp hơi mờ, đề này còn thiếu a,,bc>0
Xét \(\sqrt{\dfrac{\left(a+bc\right)\left(b+ac\right)}{c+ab}}=\sqrt{\dfrac{\left(a\left(a+b+c\right)+bc\right)\left(b\left(a+b+c\right)+ac\right)}{c\left(a+b+c\right)+ab}}\)
\(=\sqrt{\dfrac{\left(a^2+ab+ac+bc\right)\left(ab+b^2+bc+ac\right)}{ac+bc+c^2+ab}}\)
\(=\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)}{\left(a+c\right)\left(b+c\right)}}\)\(=\sqrt{\left(a+b\right)^2}=a+b\)
Tương tự cho 2 đẳng thức còn lại rồi cộng theo vế
\(P=a+b+b+c+c+a=2\left(a+b+c\right)=2\)