Ta có \(\sqrt{1+a^4}+\sqrt{1+b^4}\ge\)\(\ge\)\(\sqrt{2^2+\left(a^2+b^2\right)^2}\)(1)

Ta lại có \(\frac{a^2+b^2}{2}\ge ab\)

\(\frac{a^2+1}{2}\ge a\)

\(\frac{b^2+1}{2}\ge b\)

Từ đó => a² + b² \(\ge\)a + b + ab - 1 = \(\frac{1}{4}\)

Thế vào 1 ta được P \(\ge\)\(\frac{\sqrt{65}}{4}\)

\(\frac{9}{4}=\left(a+1\right)\left(b+1\right)\le\frac{\left(a+1\right)^2+\left(b+1\right)^2}{2}=\frac{2\left(a^2+1\right)+2\left(b^2+1\right)}{2}=a^2+b^2+2.\)

\(\Rightarrow a^2+b^2\ge\frac{1}{4}\)

\(\sqrt{1+a^4}+\sqrt{1+b^4}\ge\sqrt{\left(1+1\right)^2+\left(a^2+b^2\right)^2}\ge\sqrt{4+\left(\frac{1}{4}\right)^2}=\frac{\sqrt{17}}{2}\)

mincopxki nhé chứng minh trên cơ sở của bunhia và dấu bằng của nó cũng là bunhia

Ta có: \(a+b+c+\sqrt{abc}=4\)

\(\Rightarrow4a+4b+4c+4\sqrt{abc}=16\)

\(\Rightarrow4a+4\sqrt{abc}=16-4b-4c\)

\(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(16-4b-4c+bc\right)}=\sqrt{a\left(4a+4\sqrt{abc}+bc\right)}\)

\(=\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}=\left|2a+\sqrt{abc}\right|=2a+\sqrt{abc}\)

Tương tự: 

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{b\left(4-a\right)\left(4-c\right)}=2b+\sqrt{abc}\\\sqrt{c\left(4-a\right)\left(4-b\right)}=2c+\sqrt{abc}\end{matrix}\right.\)

\(\Rightarrow A=\sqrt{a\left(4-b\right)\left(4-c\right)}+\sqrt{b\left(4-c\right)\left(4-a\right)}+\sqrt{c\left(4-a\right)\left(4-b\right)}-\sqrt{abc}=2a+2b+2c+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c+\sqrt{abc}\right)=8\)

Ta có \(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(a+c+\sqrt{abc}\right)\left(4-c\right)}\)

\(=\sqrt{\left(a^2+ac+a\sqrt{abc}\right)\left(4-c\right)}\\ =\sqrt{4a^2+ac\left(4-\sqrt{abc}-a-c\right)+4a\sqrt{abc}}\\ =\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}\\ =2a+\sqrt{abc}\left(a,b,c>0\right)\)

Cmtt \(\sqrt{b\left(4-c\right)\left(4-a\right)}=2b+\sqrt{abc};\sqrt{c\left(4-b\right)\left(4-a\right)}=2c+\sqrt{abc}\)

\(\Rightarrow A=2\left(a+b+c\right)+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c\right)+2\sqrt{abc}\\ A=2\left(a+b+c+\sqrt{abc}\right)=2\cdot4=8\)

\(=\left(3\sqrt{7}-4\right).\sqrt{7}+7\sqrt{7}=3\sqrt{7}+3\sqrt{7}=6\sqrt{7}\)

\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

\(a+b+c+\sqrt{abc}=4\Rightarrow4a+4b+4c+4\sqrt{abc}=16\Rightarrow16-4b-4c=4a+4\sqrt{abc}\)

\(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(16-4b-4c+bc\right)}=\sqrt{a\left(4a+4\sqrt{abc}+bc\right)}\)

\(=\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}=2a+\sqrt{abc}\)

Tương tự : \(\sqrt{b\left(4-a\right)\left(4-c\right)}=2b+\sqrt{abc}\); \(\sqrt{c\left(4-a\right)\left(4-b\right)}=2c+\sqrt{abc}\)

\(\Rightarrow A=2a+2b+2c+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c+\sqrt{abc}\right)=8\)

neu de bai bai 1 la tinh x+y thi mik lam cho

đăng từng này thì ai làm cho 

We have \(P=\frac{x^4+2x^2+2}{x^2+1}\)

\(\Rightarrow P=\frac{x^4+2x^2+1+1}{x^2+1}\)

\(=\frac{\left(x^2+1\right)^2+1}{x^2+1}\)

\(=\left(x^2+1\right)+\frac{1}{x^2+1}\)

\(\ge2\sqrt{\frac{x^2+1}{x^2+1}}=2\)

(Dấu "="\(\Leftrightarrow x=0\))

Vậy \(P_{min}=2\Leftrightarrow x=0\)

Nguyễn Bùi Đại Hiệp phục bạn này lần nào hỏi cũng chép sai đề.

\(a+b+c+\sqrt{abc}=4\)

\(\Leftrightarrow4\left(a+b+c\right)+4\sqrt{abc}=16\)(*)

\(A=\Sigma\left(\sqrt{a\left(4-b\right)\left(4-c\right)}\right)-\sqrt{abc}\)

\(A=\Sigma\left(\sqrt{a\left(16-4b-4c+bc\right)}\right)-\sqrt{abc}\)

Thay (*) vào A ta được :

\(A=\Sigma\left(\sqrt{a\left(4a+4b+4c+4\sqrt{abc}-4b-4c+bc\right)}\right)-\sqrt{abc}\)

\(A=\Sigma\left(\sqrt{a\left(4a+4\sqrt{abc}+bc\right)}\right)-\sqrt{abc}\)

\(A=\Sigma\sqrt{a\left(2\sqrt{a}+\sqrt{bc}\right)^2}-\sqrt{abc}\)

\(A=\Sigma\left[\sqrt{a}\cdot\left(2\sqrt{a}+\sqrt{bc}\right)\right]-\sqrt{abc}\)

\(A=\Sigma\left(2a+\sqrt{abc}\right)-\sqrt{abc}\)

\(A=2\left(a+b+c\right)+3\sqrt{abc}-\sqrt{abc}\)

\(A=2\left(a+b+c\right)+2\sqrt{abc}\)

\(A=2\left(a+b+c+\sqrt{abc}\right)\)

\(A=2\cdot4=8\)

Vậy....

ta có \(a+b+c+\sqrt{abc}=4\Rightarrow4a+4b+4a+4\sqrt{abc}\)

=> \(4a+4\sqrt{abc}=16-4b-4c\Leftrightarrow4a+4\sqrt{abc}+bc=16-4b-4c+bc\)

=> \(\left(2\sqrt{a}+\sqrt{bc}\right)^2=\left(4-b\right)\left(4-c\right)\Rightarrow a\left(4-b\right)\left(4-c\right)=a\left(2\sqrt{a}+\sqrt{bc}\right)^2\)

=> \(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a}\left(2\sqrt{a}+\sqrt{bc}\right)=2a+\sqrt{abc}\)

tương tự như thế thay vào , thì A=8

Ta có:

\(a+b+c+\sqrt{abc}=4\Rightarrow4a+4b+4c+4\sqrt{abc}\)

\(\Rightarrow4a+4\sqrt{abc}=16-4b-4c\Leftrightarrow4a+4\sqrt{abc}+bc=16-4b-4c+bc\)

\(\Rightarrow\left(2\sqrt{a}+\sqrt{bc}\right)^2=\left(4-b\right)\left(4-c\right)\Rightarrow a\left(4-b\right)\left(4-c\right)=a\left(2\sqrt{a}+\sqrt{bc}\right)^2\)

\(\Rightarrow\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a}\left(2\sqrt{a}+\sqrt{bc}\right)=2a+\sqrt{abc}\)

Tương tự như thế thay vào, thì A = 8