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Sang học 24 tìm ai tên Perfect Blue nhé t làm bên đó rồi đưa link thì lỗi ==" , tìm tên đăng nhập springtime ấy
Ta co:
\(\sqrt{2\left(b+1\right)}\le\frac{b+3}{2}\Rightarrow\frac{a}{\sqrt{2\left(b+1\right)}}\ge\frac{2a}{b+3}\)
Tuong tu:\(\frac{b}{\sqrt{2\left(c+1\right)}}\ge\frac{2b}{c+3};\frac{c}{\sqrt{2\left(a+1\right)}}\ge\frac{2c}{a+3}\)
\(\Rightarrow\frac{1}{\sqrt{2}}\left(\frac{a}{\sqrt{b+1}}+\frac{b}{\sqrt{c+1}}+\frac{c}{\sqrt{a+1}}\right)\ge2\left(\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{a+3}\right)\)
\(\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{a+3}\)
\(=\frac{a^2}{ab+3a}+\frac{b^2}{bc+3b}+\frac{c^2}{ca+3c}\ge\frac{\left(a+b+c\right)^2}{ab+bc+ca+9}\ge\frac{\left(a+b+c\right)^2}{\frac{\left(a+b+c\right)^2}{3}+9}=\frac{9}{\frac{9}{3}+9}=\frac{3}{4}\)
\(\Rightarrow2\left(\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{a+3}\right)\ge\frac{3}{2}\)
Hay \(\frac{a}{\sqrt{b+1}}+\frac{b}{\sqrt{c+1}}+\frac{c}{\sqrt{a+1}}\ge\frac{3\sqrt{2}}{2}\)
Dau '=' xay ra khi \(a=b=c=3\)
Thắng Nguyễn Phần cuối cùng viết rõ ra một chút :
\(2\sqrt{2}\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)\ge\frac{y^2+z^2-x^2}{x}+\frac{y^2+x^2-z^2}{z}+\frac{x^2+z^2-y^2}{y}\)
\(\frac{y^2}{x}+\frac{z^2}{x}+\frac{y^2}{z}+\frac{x^2}{z}+\frac{x^2}{y}+\frac{z^2}{y}-\sqrt{2015}\ge\frac{\left[2\left(x+y+z\right)\right]^2}{2\left(x+y+z\right)}-\sqrt{2015}=\sqrt{2015}\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\sqrt{2015}}{2\sqrt{2}}=\frac{1}{2}\sqrt{\frac{2015}{2}}\)
Đặt \(\sqrt{a^2+b^2=z};\sqrt{a^2+c^2}=y;\sqrt{b^2+c^2}=x\left(x;y;z>0\right)\)
\(\Rightarrow a^2=\frac{y^2+z^2-x^2}{2};b=\frac{x^2+z^2-y^2}{2};c=\frac{x^2+y^2-z^2}{2}\)
Theo đề \(x+y+z=\sqrt{2015}\)
Ta có:\(b+c\le\sqrt{2\left(b^2+c^2\right)}=\sqrt{2}\cdot x\)\(\Rightarrow\frac{a^2}{b+c}\ge\frac{y^2+z^2-x^2}{2\sqrt{2}\cdot x}\)
Tương tự cho 2 cái còn lại rồi, cộng lại:
\(VT\cdot2\sqrt{2}\ge\sqrt{2015}\Rightarrow VT\ge\frac{1}{2}\sqrt{\frac{2015}{2}}\)
ơ đang chờ mấy bạn top bxh vô trả lời mà hỏng thấy đou
hộ mình với:(
\(a^2\sqrt{a}+b^2\sqrt{b}+c^2\sqrt{c}+\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)
\(=\left(a^2\sqrt{a}+\frac{1}{\sqrt{a}}\right)+\left(b^2\sqrt{b}+\frac{1}{\sqrt{b}}\right)+\left(c^2\sqrt{c}+\frac{1}{\sqrt{c}}\right)\)
\(\ge2a+2b+2c\ge6\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=6\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)
Từ ( 1 ) và ( 2 ) có đpcm
1,
\(\frac{a}{1+\frac{b}{a}}+\frac{b}{1+\frac{c}{b}}+\frac{c}{1+\frac{a}{c}}=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\ge\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\frac{2}{2}=1\left(Q.E.D\right)\)
Đặt \(\hept{\begin{cases}\sqrt{a^2+b^2}=x\\\sqrt{b^2+c^2}=y\\\sqrt{c^2+a^2}=z\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x,y,z>0\\x+y+z=1\end{cases}}\)
Và \(\hept{\begin{cases}a^2=\frac{x^2+z^2-y^2}{2}\\b^2=\frac{x^2+y^2-z^2}{2}\\c^2=\frac{y^2+z^2-x^2}{2}\end{cases}}\) và \(\hept{\begin{cases}b+c\le\sqrt{2\left(b^2+c^2\right)}=\sqrt{2}y\\a+b\le\sqrt{2}x\\c+a\le\sqrt{2}z\end{cases}}\)
\(\Rightarrow VT\ge\frac{1}{2\sqrt{2}}\left(\frac{x^2+z^2-y^2}{y}+\frac{x^2+y^2-z^2}{2z}+\frac{y^2+z^2-x^2}{x}\right)\)
\(\ge\frac{1}{2\sqrt{2}}\left(\frac{2\left(x+y+z\right)^2}{x+y+z}-\left(x+y+z\right)\right)\)
\(=\frac{1}{2\sqrt{2}}\left(x+y+z\right)=\frac{1}{2\sqrt{2}}\)