Ta có bđt quen thuộc sau \(\frac{x}{y+z}< \frac{x+m}{y+z+m}\) 

Áp dụng ta được \(\frac{a}{b+c}< \frac{a+a}{a+b+c}=\frac{2a}{a+b+c}\)
Chứng minh tương tự \(\frac{b}{c+a}< \frac{2b}{a+b+c}\)

                                     \(\frac{c}{a+b}< \frac{2c}{a+b+c}\)

Do đó \(VT< \frac{2a+2b+2c}{a+b+c}=2\)

Ta đi chứng minh VP > 2 

Áp dụng bđt Cô-si có \(a+\left(b+c\right)\ge2\sqrt{a\left(b+c\right)}\)

                             \(\Rightarrow\sqrt{a\left(b+c\right)}\le\frac{a+b+c}{2}\)

                             \(\Rightarrow\sqrt{\frac{b+c}{a}}\le\frac{a+b+c}{2a}\)

                             \(\Rightarrow\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\)

Chứng minh tương tự \(\sqrt{\frac{b}{c+a}}\ge\frac{2b}{a+b+c}\)

                                    \(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\)

Cộng 3 vế lại ta được \(VP\ge\frac{2a+2b+2c}{a+b+c}=2\)

Do đó \(VP\ge2>VT\)

\(\Rightarrow VT< VP\left(Q.E.D\right)\)

Dấu "=" không xảy ra

a=8 vì 8 + 1 = 9 = 3^2

3a + 1 =24 +1 = 25 = 5^2 

suy ra a=8

Tính :\(a,\)\(-\sqrt{\left(-6\right)^2}=-|-6|=-6\)

\(b,\)\(-\sqrt{\frac{-25}{-16}}=-\sqrt{\left(\frac{5}{4}\right)^2}=-|\frac{5}{4}|=-\frac{5}{4}\)

\(c,\)\(\sqrt{-\frac{-9}{25}}=\sqrt{\frac{9}{25}}=\sqrt{\left(\frac{3}{5}\right)^2}=|\frac{3}{5}|=\frac{3}{5}\)

\(d,\)\(\left(-\sqrt{7}\right)^2=7\)

\(e,\)\(-\left(\frac{\sqrt{3}}{4}\right)^2=-\frac{\sqrt{3}^2}{4^2}=-\frac{3}{16}\)

\(f,\)\(\sqrt{\left(-2\right)^4}=\sqrt{\left[\left(-2\right)^2\right]^2}=|-2^2|=4\)

So sánh :\(a,\) \(\sqrt{8}-1\)

\(2=3-1=\sqrt{9}-1\)

\(\Rightarrow\sqrt{8}-1< 2\)

\(b,\)\(\sqrt{\frac{16}{2}}=\sqrt{8}>\sqrt{3}\)

\(\Rightarrow\sqrt{\frac{16}{2}}>\sqrt{3}\)

\(x+y+z=a\Rightarrow\left(x+y+z\right)^2=a^2\Rightarrow xy+yz+zx=\frac{a^2-b}{2}\\ \)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\Rightarrow\frac{xy+yz+xz}{xyz}=\frac{1}{c}\Rightarrow xyz=\frac{\left(a^2-b\right)c}{2}\)

Ta có

\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)

\(=\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-\left(xy+yz+zx\right)\right)\)

\(\Rightarrow x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-\left(xy+yz+xz\right)\right)+3xyz\)

\(=a\left(b-\frac{a^2-b}{2}\right)+3\frac{\left(a^2-b\right)c}{2}\)

1/ Bình phương hai vế, ta cần chứng minh \(a+b+2\sqrt{ab}>a+b\Leftrightarrow2\sqrt{ab}>0\)

Mà ta có \(2\sqrt{ab}\ge0\text{ Nhưng theo đề bài dấu "=" không xảy ra nên ta có đpcm. }\)

Ta có: \(VT=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\frac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}=VP\)