thu gọn
B=\(a+\sqrt{a^2+\frac{\left(1+b^2\right)^2-4b^2}{\left(b^2+1\right)^2}}\)
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\(=\left(\frac{\sqrt{1+m}}{\sqrt{1+m}-\sqrt{1-m}}+\frac{\sqrt{1-m}\cdot\sqrt{1-m}}{\sqrt{1-m}\cdot\left(\sqrt{1+m}-\sqrt{1-m}\right)}\right)\cdot\frac{\sqrt{1-m^2}-1}{m}\)
\(=\frac{\sqrt{1+m}+\sqrt{1-m}}{\sqrt{1+m}-\sqrt{1-m}}\cdot\frac{\sqrt{1-m^2}-1}{m}\)
\(=\frac{\left(\sqrt{1+m}+\sqrt{1-m}\right)^2}{\left(\sqrt{1+m}-\sqrt{1-m}\right)\left(\sqrt{1+m}+\sqrt{1-m}\right)}\cdot\frac{\sqrt{1-m^2}-1}{m}\)
\(=\frac{1+m-m+1+2\sqrt{1-m^2}}{2m}\cdot\frac{\sqrt{1-m^2}-1}{m}\)
\(=\frac{\sqrt{1-m^2}+1}{m}\cdot\frac{\sqrt{1-m^2}-1}{m}=\frac{1-m^2-1}{m^2}=-1\)
1)
Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)
Dấu "=" xảy ra khi a=b=c
2)
\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)
Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)
\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)
Áp dụng BĐT bunniacoxki ta có:
\(\left(b^2+\left(c+a\right)^2\right)\left(1+4\right)\ge\left(b+2\left(a+c\right)\right)^2\)
=> \(\sqrt{\frac{a^2}{b^2+\left(c+a\right)^2}}\le\sqrt{5}.\frac{a}{b+2c+2a}\)
=> \(VT\le\sqrt{5}.\left(\frac{a}{b+2c+2a}+\frac{b}{c+2a+2b}+\frac{c}{a+2b+2c}\right)\)
Cần CM \(\frac{a}{b+2c+2a}+\frac{b}{c+2a+2b}+\frac{c}{a+2b+2c}\le\frac{3}{5}\)
<=>\(\left(\frac{1}{2}-\frac{a}{b+2c+2a}\right)+\left(\frac{1}{2}-\frac{b}{c+2a+2b}\right)+\left(\frac{1}{2}-\frac{c}{a+2b+2c}\right)\ge\frac{9}{10}\)
<=>\(\frac{b+2c}{b+2c+2a}+\frac{c+2a}{c+2a+2b}+\frac{a+2b}{a+2b+2c}\ge\frac{9}{5}\)
Áp dụng bđt buniacoxki dạng phân thức ở vế trái:
=> \(VT\ge\frac{\left(b+2c+c+2a+a+2b\right)^2}{\left(b+2c\right)^2+2a\left(b+2c\right)+\left(c+2a\right)^2+2b\left(c+2a\right)+\left(a+2b\right)^2+2c\left(a+2b\right)}\)
\(=\frac{9\left(a+b+c\right)^2}{5\left(a+b+c\right)^2}=\frac{9}{5}\)(ĐPCM)
Dấu bằng xảy ra khi a=b=c
\(1,\sqrt{4\left(a-4\right)^2}\left(dkxd:a\ge4\right)\)
\(=\sqrt{4}.\sqrt{\left(a-4\right)^2}\)
\(=\sqrt{2^2}.\left|a-4\right|\)
\(=2\left(a-4\right)\)
\(=2a-8\)
\(2,\sqrt{9\left(b-5\right)^2}\left(dkxd:b< 5\right)\)
\(=\sqrt{9}.\sqrt{\left(b-5\right)^2}\)
\(=\sqrt{3^2}.\left|b-5\right|\)
\(=3\left(-b+5\right)\)
\(=-3b+15\)
\(a,\left(1+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(1-\frac{a+\sqrt{a}}{1+\sqrt{a}}\right)=\left(1+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1^2-\sqrt{a}^2=1-a\)
\(b,\left(2-\frac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\frac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)=\left(2-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2-\frac{-\sqrt{a}\left(\sqrt{b}-5\right)}{\sqrt{b}-5}\right)\)
\(=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)
\(c,\left(3+\frac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3-\frac{3a+\sqrt{a}}{3\sqrt{a}+1}\right)=\left(3+\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\right)\left(3-\frac{\sqrt{a}\left(3\sqrt{a}+1\right)}{3\sqrt{a}+1}\right)\)
\(=\left(3+\sqrt{a}\right)\left(3-\sqrt{a}\right)=3^2-\sqrt{a}^2=3-a\)
\(d,\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}+a}{1+\sqrt{a}}\right)=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(\sqrt{a}+2\right)\left(2-\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)
B=a+\(\sqrt{\frac{\left(1+b^2-2b\right)\left(1+b^2+2b\right)}{\left(b^2+1\right)^2}+a^2}\)=a+\(\sqrt{\frac{\left(b^2-1\right)^2}{\left(b^2+1\right)^2}+a^2}\)