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\(\sqrt{4\left(a-3\right)^2}\)
\(=\sqrt{2^2\left(a-3\right)^2}\)
\(=2\left(a-3\right)\)
\(=2a-6\)
\(\sqrt{4\left(a-3\right)^2}=\sqrt{\left[2\left(a-3\right)\right]^2}=2\left(a-3\right)\)3)
a) \(\sqrt{\left(3-6a\right)^2}=6a-3\)
( vì \(a\ge\frac{1}{2}\)\(\Rightarrow3-6a< 0\))
ta có:\(\sqrt{\frac{b+c}{a}}\le\frac{a+b+c}{2a}.\) (BĐT cauchy)
\(\Rightarrow\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\) (1)
tương tự ta có: \(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c}\) (2)
\(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\) (3)
từ (1),(2),(3) => \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2\left(a+b+c\right)}{a+b+c}=2\)
=> đpcm
ta có ; \(\sqrt{\frac{a}{b+c}}=\frac{a}{\sqrt{a\left(b+c\right)}}\ge\frac{2a}{a+b+c}\)
\(\sqrt{\frac{b}{c+a}}=\frac{b}{\sqrt{b\left(c+a\right)}}\ge\frac{2b}{a+b+c}\)
\(\sqrt{\frac{c}{a+b}}=\frac{c}{\sqrt{c\left(a+b\right)}}\ge\frac{2c}{a+b+c}\)
cộng lại theo từng vế ta có biểu thức đó \(\ge2\). xảy ra đẳng thức \(\hept{\begin{cases}a=b+c\\b=a+c\\c=a+b\end{cases}\Rightarrow a+b+c=0\left(\ne gt\right)}\)
\(\Rightarrow\)đẳng thức ko xảy ra
1) \(VT=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}=VP\)(ĐPCM)
2) \(VT=\text{[}\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a+b-\sqrt{ab}\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\text{]}.\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)
\(=\frac{\left(a+b-\sqrt{ab}-\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}=\frac{\left(a-b\right)^2}{\left(a-b\right)^2}=1=VP\)(ĐPCM)
4) \(VT=\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a=VP\)(ĐPCM)
