2ab =a^2+b^2+2ab

=>a^2+b^2=0

Mà a^2 >=0 ;b^2>=0 => a^2+b^2>=0 

=> a^2= b^2=0

=> a=b=0

Ta có : ab + ab = ( a + b )²

\(\Leftrightarrow\)( a + b )² - ( ab + ab )= 0

\(\Leftrightarrow\)a² + 2ab + b² - 2ab= 0

\(\Leftrightarrow\)( a² + b² ) + ( 2ab - 2ab ) = 0

\(\Leftrightarrow\)a² + b² = 0

\(\Leftrightarrow\)a² = 0 và b² = 0

\(\Leftrightarrow\)a = 0 và b = 0

Vậy : a = b = 0

Lời giải

Ta có: \(\left(a+b+\frac{1}{4}\right)^2=\frac{1}{16}\left(4a+4b-1\right)^2+\left(a+b\right)\ge a+b\)

Tương tự: \(\left(b+c+\frac{1}{4}\right)^2\ge b+c;\left(c+a+\frac{1}{4}\right)^2\ge c+a\)

Như vậy: \(L.H.S\left(VT\right)\ge\left(a+b\right)+\left(b+c\right)+\left(c+a\right)=\left(\frac{1}{\frac{1}{a}}+\frac{1}{\frac{1}{b}}\right)+\left(\frac{1}{\frac{1}{b}}+\frac{1}{\frac{1}{c}}\right)+\left(\frac{1}{\frac{1}{c}}+\frac{1}{\frac{1}{a}}\right)\)

\(\ge4\left(\frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{b}+\frac{1}{c}}+\frac{1}{\frac{1}{c}+\frac{1}{a}}\right)=R.H.S\left(VP\right)\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{8}\). Ta có đpcm.

khác cách tth xíu

Ta có:

\(VP=\Sigma_{cyc}\frac{4}{\frac{1}{a}+\frac{1}{b}}\le\Sigma_{cyc}\frac{4}{\frac{4}{a+b}}=2\left(a+b+c\right)\)

Gio ta di chung minh

\(VT\ge2\left(a+b+c\right)\)

Ta lai co:

\(VT=\Sigma_{cyc}\left(a+b+\frac{1}{4}\right)^2\ge\frac{\left[2\left(a+b+c\right)+\frac{3}{4}\right]^2}{3}\)

Chung minh

\(\frac{\left[2\left(a+b+c\right)+\frac{3}{4}\right]^2}{3}\ge2\left(a+b+c\right)\)

\(\Leftrightarrow\left[2\left(a+b+c\right)-\frac{3}{4}\right]^2\ge0\) (đúng)

Dau '=' xay ra khi \(a=b=c=\frac{1}{8}\)

\(a)\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)

\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2+\left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)

\(=\frac{7+2\sqrt{35}+5+7-2\sqrt{35}+5}{7-5}\)

\(=\frac{24}{2}\)

\(=12\)

\(b)\frac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)

\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)+\left(2+\sqrt{8}-\sqrt{6}\right)}{2+\sqrt{2}-\sqrt{3}}\)

\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)+\sqrt{2}\left(\sqrt{2}+2-\sqrt{3}\right)}{2+\sqrt{2}-\sqrt{3}}\)

\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}\right)}{2+\sqrt{2}-\sqrt{3}}\)

\(=1+\sqrt{2}\)

\(c)A=\left(\sqrt{3}+1\right)\sqrt{\frac{14-6\sqrt{3}}{5+\sqrt{3}}}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{70-14\sqrt{3}-30\sqrt{3}+18}{25-3}}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{88-44\sqrt{3}}{22}}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{44\left(2-\sqrt{3}\right)}{22}}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{2\left(2-\sqrt{3}\right)}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)

\(A=3-1=2\)

P/s: nếu đề là vậy thì t ra kết quả như vậy ạ, nhưng lần sau khi đăng câu hỏi bạn nên viết rõ hơn ra nhé

\(\left(a^2+b^2+c^2\right)^2\ge a^4+b^4+c^4+a^2b^2+b^2c^2+c^2a^2\)

\(\ge a^4+b^4+c^4+a^2b^2-2abc^2\)

\(=\left(a^2+b^2+c^2\right)\left(a^4+b^4+\left(c^2-ab\right)^2\right)\)

\(\ge\left(a^3+b^3+c\left(c^2-ab\right)\right)^2\)

\(=\left(a^3+b^3+c^3-abc\right)^2\ge\left(a^3+b^3+c^3-3abc\right)^2=1\)

\(\Rightarrow B\ge1\)

Câu hỏi của Giao Khánh Linh - Toán lớp 9 - Học toán với OnlineMath