Áp dụng Bu-nhi-a-cốp-xki:

(\( {4 \over 9}\)+\( {3 \over 2}\)) (a² +\( {1 \over b^2}\))   >=  \({2a \over 3}\) +\( { \sqrt{3}b \over 2}\)

\(VT\ge\sqrt{\left(a+b+c\right)^2+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}\)

\(VT\ge\sqrt{\left(a+b+c\right)^2+\frac{81}{\left(a+b+c\right)^2}}\)

\(VT\ge\sqrt{\left(a+b+c\right)^2+\frac{16}{\left(a+b+c\right)^2}+\frac{65}{\left(a+b+c\right)^2}}\)

\(VT\ge\sqrt{2\sqrt{\frac{16\left(a+b+c\right)^2}{\left(a+b+c\right)^2}}+\frac{65}{2^2}}=\frac{\sqrt{97}}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{2}{3}\)

bạn viết sai đề rồi nhé đề đúng là căn(b^2+1/c^2) và căn (c^2 + 1/a^2) ở vế trái chứ ?

Áp dụng BĐT Cô - si, ta có :

\(\left(1.a+\frac{9}{4}.\frac{1}{b}\right)^2\le\left(1^2+\frac{81}{16}\right)\left(a^2+\frac{1}{b^2}\right)\)

\(\Rightarrow\sqrt{a^2+\frac{1}{b^2}}\ge\frac{4}{\sqrt{97}}\left(a+\frac{9}{4b}\right)\).Chứng minh tương tự, ta có:

\(\sqrt{b^2+\frac{1}{c^2}}\ge\frac{4}{\sqrt{97}}\left(b+\frac{9}{4c}\right)\)

\(\sqrt{c^2+\frac{1}{a^2}}\ge\frac{4}{\sqrt{97}}\left(c+\frac{4}{9a}\right)\)

Cộng 3 vế BĐT => đpcm

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{c}\ge4\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\ge2\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1\)

Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z\ge1\)

\(P=\sqrt{x^2+2y^2}+\sqrt{y^2+2z^2}+\sqrt{z^2+2x^2}\)

\(\Rightarrow P\ge\sqrt{\frac{\left(x+2y\right)^2}{3}}+\sqrt{\frac{\left(y+2z\right)^2}{3}}+\sqrt{\frac{\left(z+2x\right)^2}{3}}\)

\(\Rightarrow P\ge\frac{1}{\sqrt{3}}\left(3x+3y+3z\right)\ge\frac{3}{\sqrt{3}}=\sqrt{3}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\) hay \(a=b=c=3\)

dung bunicopxi di

minh moi hok lop 6 thoi

xin lỗi em mới học lớp 5 duyetj đi

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:

\(VT\cdot2\sqrt{2}\ge\sqrt{2015}\Rightarrow VT\ge\frac{1}{2}\sqrt{\frac{2015}{2}}\)

khong biet

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