Áp dụng Côsi:

\(a^2+\left(\frac{19-\sqrt{37}}{12}\right)^2\ge2\sqrt{\left(\frac{19-\sqrt{37}}{12}\right)^2.a^2}=2.\frac{19-\sqrt{37}}{12}a\)

\(b^2+\left(\frac{19-\sqrt{37}}{12}\right)^2\ge2.\frac{19-\sqrt{37}}{12}b\)

\(c^3+\left(\frac{\sqrt{37}-1}{6}\right)^3+\left(\frac{\sqrt{37}-1}{6}\right)^3\ge3\sqrt[3]{\left(\frac{\sqrt{37}-1}{6}\right)^3\left(\frac{\sqrt{37}-1}{6}\right)^3.c^3}=3.\left(\frac{\sqrt{37}-1}{6}\right)^2c\)

\(\Rightarrow a^2+b^2+c^3+2\left(\frac{19-\sqrt{37}}{12}\right)^2+2\left(\frac{\sqrt{37}-1}{6}\right)^3\ge2.\frac{19-\sqrt{37}}{12}a+2.\frac{19-\sqrt{37}}{12}b+3.\left(\frac{\sqrt{37}-1}{6}\right)^2c\)

\(\Rightarrow a^2+b^2+c^3+2.\left(\frac{19-\sqrt{37}}{12}\right)^2+3.\left(\frac{\sqrt{37}-1}{6}\right)^3\ge\frac{19-\sqrt{37}}{6}\left(a+b+c\right)=\frac{19-\sqrt{37}}{2}\)

\(\Rightarrow a^2+b^2+c^3\ge\frac{19-\sqrt{37}}{2}-2.\left(\frac{19-\sqrt{37}}{12}\right)^2-2.\left(\frac{\sqrt{37}-1}{6}\right)^3\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=\frac{19-\sqrt{37}}{12};\text{ }c=\frac{\sqrt{37}-1}{6}\)

Vậy GTNN của biệu thức là .......

Nhân P với 4. Do 4=a+b+c+d+e

Áp dụng \(\left(x+y\right)^2\ge4xy\)

Nhân 16, xin lỗi mình nhầm

@Nguyễn Việt Lâm

@Lê Thị Thục Hiền

Làm được ở TH a,b,c > hoặc = 0 thôi nha ( nếu a,b,c>0 thì mình chỉ biết tìm maxP thôi)

Đặt \(\sqrt{a+b}=x\), \(\sqrt{b+c}=y\),\(\sqrt{c+a}=z\) (x,y,z \(\ge0\))

=> \(0\le x,y,z\le2\)

Có \(x^2+y^2+z^2=2\left(a+b+c\right)=2.4=8\)

Có \(2-x\ge0\) => \(x\left(2-x\right)\ge0\) <=> \(2x-x^2\ge0\) <=> \(2x\ge x^2\)

Cm tương tự cũng có: \(2y\ge y^2\) , \(2z\ge z^2\)

=>\(2x+2y+2z\ge x^2+y^2+z^2=8\)

<=> \(x+y+z\ge4\)

<=> \(P=x+y+z\ge4\)

Dấu "=" xảy ra <=>\(\left(x,y,z\right)\in\left(2,2,0\right),\left(2,0,2\right),\left(0,2,2\right)\)

=> \(\left(a,b,c\right)\in\left\{\left(0,4,0\right),\left(4,0,0\right),\left(0,0,4\right)\right\}\)

Áp dụng BĐT AM-GM ta có:

\(A=5a+6b+7c+\frac{1}{a}+\frac{8}{b}+\frac{27}{c}\)

\(=4\left(a+b+c\right)+\left(\frac{1}{a}+a\right)+\left(\frac{8}{b}+2b\right)+\left(\frac{27}{c}+3c\right)\)

\(\ge4\cdot6+2\sqrt{\frac{1}{a}\cdot a}+2\sqrt{\frac{8}{b}\cdot2b}+2\sqrt{\frac{27}{c}\cdot3c}\)

\(\ge24+2+2\cdot4+2\cdot9=52\)

Xảy ra khi \(\frac{1}{a}=a;\frac{8}{b}=2b;\frac{27}{c}=3c\Rightarrow a=1;b=2;c=3\)

Lời giải:

Ta có: \(A=\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+4}{c}\)

\(\Leftrightarrow A=1+\frac{1}{a}+1+\frac{1}{b}+1+\frac{4}{c}=3+\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\)

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)(a+b+c)\geq (1+1+2)^2\)

\(\Leftrightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\geq \frac{4^2}{a+b+c}=\frac{16}{6}=\frac{8}{3}\)

Do đó: \(A\geq 3+\frac{8}{3}=\frac{17}{3}\) hay \(A_{\min}=\frac{17}{3}\)

Dấu bằng xảy ra khi \((a,b,c)=(\frac{3}{2}; \frac{3}{2}; 3)\)