Biến đổi tương đương:

\(\dfrac{a^2}{4}+b^2+c^2\ge ab-ac+2bc\) (1)

\(\Leftrightarrow a^2+4b^2+4c^2\ge4ab-4ac+8bc\)

\(\Leftrightarrow a^2+\left(2b\right)^2+\left(2c\right)^2-2.a.2b+2.a.2c-2.2b.2c\ge0\)

\(\Leftrightarrow\left(a-2b+2c\right)^2\ge0\) luôn đúng

=> (1) đúng

Dấu "=" xảy ra khi a = 2(b - c)

Áp dụng BĐT AM - GM, ta có:

\(\dfrac{a^3+b^3+c^3}{2abc}+\dfrac{a^2+b^2}{c^2+ab}+\dfrac{b^2+c^2}{a^2+bc}+\dfrac{a^2+c^2}{b^2+ac}\)

\(\ge\dfrac{3\sqrt{a^3b^3c^3}}{2abc}+\dfrac{a^2+b^2}{c^2+\dfrac{a^2+b^2}{2}}+\dfrac{b^2+c^2}{a^2+\dfrac{b^2+c^2}{2}}+\dfrac{a^2+c^2}{b^2+\dfrac{a^2+c^2}{2}}\)

\(\ge\dfrac{3abc}{2abc}+\dfrac{2\left(a^2+b^2\right)}{2c^2+a^2+b^2}+\dfrac{2\left(b^2+c^2\right)}{2a^2+b^2+c^2}+\dfrac{2\left(a^2+c^2\right)}{2b^2+a^2+c^2}\)

\(=\dfrac{3}{2}+2\times\left[\dfrac{a^2+b^2}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}+\dfrac{b^2+c^2}{\left(a^2+b^2\right)+\left(a^2+c^2\right)}+\dfrac{c^2+a^2}{\left(b^2+c^2\right)+\left(b^2+a^2\right)}\right]\) (1)

Đặt \(\left\{{}\begin{matrix}a^2+b^2=x\\b^2+c^2=y\\c^2+a^2=z\end{matrix}\right.\), ta có:

\(\left(1\right)\Leftrightarrow\dfrac{3}{2}+2\times\left(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\right)\)

\(\ge\dfrac{3}{2}+2\times\dfrac{3}{2}\) (Bất_đẳng_thức_Nesbitt)

\(=\dfrac{9}{2}\left(\text{đ}pcm\right)\)

Dấu "=" xảy ra khi a = b = c

Đặt \(A=\dfrac{\sqrt{\sqrt{5}+2}-\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)

\(\Rightarrow A^2=\dfrac{\left(\sqrt{5}+2\right)-2\sqrt{5-4}+\left(\sqrt{5}-2\right)}{\sqrt{5}+1}\)

\(=\dfrac{2\sqrt{5}-2}{\sqrt{5}+1}\)

\(=\dfrac{2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}{5-1}\)

\(=\dfrac{2\left(5-1\right)}{5-1}\)

\(=2\)

\(\Rightarrow A=\sqrt{2}\left(A>0\right)\)

\(D=A-\sqrt{3-2\sqrt{2}}\)

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

\(=1\)

Áp dụng BĐT Cauchy Shwarz, ta có:

\(\left(2^2+1^2\right)\left[\left(3x\right)^2+y^2\right]\ge\left(6x+y\right)^2\)

\(\Leftrightarrow9x^2+y^2\ge5\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=1\end{matrix}\right.\)

Áp dụng BĐT Cauchy Shwarz, ta có:

\(\left(a+b\right)^2\le2\left(a^2+b^2\right)=2\)

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

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

\(a^2-b=b^2-c\)

\(\Leftrightarrow a^2-b^2=b-c\)

\(\Leftrightarrow b-c=\left(a-b\right)\left(a+b\right)\)

\(\Leftrightarrow\dfrac{a+b}{b-c}=\dfrac{1}{a-b}=\dfrac{a+b+1}{b-c+a-b}=\dfrac{a+b+1}{a-c}\)

\(\Leftrightarrow a+b+1=\dfrac{a-c}{a-b}\)

Tương tự, ta có: \(a+c+1=\dfrac{b-c}{a-c}\) và \(b+c+1=\dfrac{b-a}{b-c}\)

Nhân vế theo vế, ta có P = - 1.

\(\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}\)

\(=\sqrt{1^2+a^2+\left(\dfrac{a}{a+1}\right)^2+2a-\dfrac{2a}{a+1}-\dfrac{2a^2}{a+1}}\)

(vì \(2a-\dfrac{2a}{a+1}-\dfrac{2a^2}{a+1}=\dfrac{2a^2+2a-2a-2a^2}{a+1}=0\))

\(=\sqrt{\left(1+a-\dfrac{a}{a+1}\right)^2}\)

\(=\left|1+a-\dfrac{a}{a+1}\right|\)

Áp dụng vào P, ta có:

\(P=\sqrt{1+1999^2+\dfrac{1999^2}{2000^2}}+\dfrac{1999}{2000}\)

\(=\left|1+1999-\dfrac{1999}{2000}\right|+\dfrac{1999}{2000}\)

\(=2000\)

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

\(\Leftrightarrow\sqrt{x}+\left(\sqrt{1-x}-1\right)+\left(\sqrt{1+x}-1\right)=0\)

\(\Leftrightarrow\sqrt{x}+\dfrac{1-x-1}{\sqrt{1-x}+1}+\dfrac{1+x-1}{\sqrt{1+x}+1}=0\)

\(\Leftrightarrow\sqrt{x}-\dfrac{x}{\sqrt{1-x}+1}+\dfrac{x}{\sqrt{1+x}+1}=0\)

\(\Leftrightarrow\sqrt{x}\left(1-\dfrac{\sqrt{x}}{\sqrt{1-x}+1}+\dfrac{\sqrt{x}}{\sqrt{1+x}+1}\right)=0\)

Pt \(1-\dfrac{\sqrt{x}}{\sqrt{1-x}+1}+\dfrac{\sqrt{x}}{\sqrt{1+x}+1}=0\) vô n_o

=> x = 0 (nhận)

\(a=\sqrt{5}+\sqrt{\dfrac{1}{5}}=\sqrt{5}+\dfrac{1}{\sqrt{5}}=\dfrac{5+1}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)

\(\sqrt{5a^2-4a\sqrt{5}+4}\)

\(=\sqrt{\left(\sqrt{5}a\right)^2-4\times\left(\sqrt{5}a\right)+4}\)

\(=\sqrt{\left(\sqrt{5}a-2\right)^2}\)

\(=\left|\sqrt{5}a-2\right|\)

\(=\left|\sqrt{5}\times\dfrac{6\sqrt{5}}{5}-2\right|\)

= 4

~ ~ ~

\(\dfrac{\sqrt{1-2m+m^2}}{m^2-1}\)

\(=\dfrac{\sqrt{\left(1-m\right)^2}}{\left(m-1\right)\left(m+1\right)}\)

\(=\dfrac{\left|1-m\right|}{\left(m-1\right)\left(m+1\right)}\)

\(=\dfrac{m-1}{\left(m-1\right)\left(m+1\right)}\) (m > 1)

\(=\dfrac{1}{m+1}\)

\(\sqrt{3+2\sqrt{x}}+\sqrt{x}=6\)

\(\Leftrightarrow\left(\sqrt{3+2\sqrt{x}}-3\right)+\left(\sqrt{x}-3\right)=0\)

\(\Leftrightarrow\dfrac{3+2\sqrt{x}-9}{\sqrt{3+2\sqrt{x}}+3}+\left(\sqrt{x}-3\right)=0\)

\(\Leftrightarrow\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{3+2\sqrt{x}}+3}+\left(\sqrt{x}-3\right)=0\)

\(\Leftrightarrow\left(\dfrac{2}{\sqrt{3+2\sqrt{x}}+3}+1\right)\left(\sqrt{x}-3\right)=0\)

Pt \(\dfrac{2}{\sqrt{3+2\sqrt{x}}+3}+1=0\) vô n_o

\(\Rightarrow\sqrt{x}-3=0\)

\(\Leftrightarrow x=9\) (nhận)