Áp dụng BĐT BSC:

\(A=\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac{1}{z}\)

\(=\dfrac{\dfrac{1}{16}}{x}+\dfrac{\dfrac{1}{4}}{y}+\dfrac{1}{z}\)

\(\ge\dfrac{\left(\dfrac{1}{4}+\dfrac{1}{2}+1\right)^2}{x+y+z}=\dfrac{49}{16}\)

\(minA=\dfrac{49}{16}\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\dfrac{1}{4}}{x}=\dfrac{\dfrac{1}{2}}{y}=\dfrac{1}{z}\\x+y+z=1\end{matrix}\right.\)

\(\Leftrightarrow\left(x;y;z\right)=\left(\dfrac{1}{7};\dfrac{2}{7};\dfrac{4}{7}\right)\)

\(P=\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac{1}{z}+\dfrac{49}{16}-\dfrac{49}{16}\)

\(P=\left(\dfrac{1}{16x}+\dfrac{49x}{16}\right)+\left(\dfrac{1}{4y}+\dfrac{49y}{16}\right)+\left(\dfrac{1}{z}+\dfrac{49z}{16}\right)-\dfrac{49}{16}\)

\(P\ge2\sqrt{\dfrac{49x}{16x.16}}+2\sqrt{\dfrac{49y}{4y.16}}+2\sqrt{\dfrac{49z}{z.16}}-\dfrac{49}{16}=\dfrac{49}{16}\)

Dấu "=" xảy ra khi...

Ta có: \(\sqrt{\left(x^2+\dfrac{1}{y^2}\right)\left(1+81\right)}\ge\sqrt{\left(x+\dfrac{9}{y}\right)^2}\)

=> \(\sqrt{x^2+\dfrac{1}{y^2}}\ge\dfrac{x+\dfrac{9}{y}}{\sqrt{82}}\)

Tương tự => \(\left\{{}\begin{matrix}\sqrt{y^2+\dfrac{1}{z^2}}\ge\dfrac{y+\dfrac{9}{z}}{\sqrt{82}}\\\sqrt{z^2+\dfrac{1}{x^2}}\ge\dfrac{z+\dfrac{9}{x}}{\sqrt{82}}\end{matrix}\right.\)

=> \(P\ge\dfrac{\left(x+y+z\right)+9\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}{\sqrt{82}}\)

Mà x + y + z = 1

      \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}=9\)

=> \(P\ge\sqrt{82}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{3}\)

\(S=\left(\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}\right)=\left(\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}\right).\left(x+y+z\right)\)   (do x+y+z=1 nên michf nhân vào kết quả sẽ ko bị  thay đổi)

\(S=\frac{21}{16}+\left(\frac{x}{4y}+\frac{y}{16x}\right)+\left(\frac{x}{z}+\frac{z}{16x}\right)+\left(\frac{y}{z}+\frac{z}{4y}\right)\)

AD BĐT cô si,ta có:

\(S\ge\frac{21}{16}+2.\sqrt{\frac{x}{4y}.\frac{y}{16x}}+2\sqrt{\frac{x}{z}.\frac{z}{16x}}+2.\sqrt{\frac{y}{z}.\frac{z}{4y}}=\frac{21}{16}+\frac{1}{4}+\frac{1}{2}+1=\frac{49}{16}\)

dấu bằng xảy ra \(\Leftrightarrow\hept{\begin{cases}4x=2y=z\\x+y+z=1\\x;y;z>0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{7}\\y=\frac{2}{7}\\z=\frac{4}{7}\end{cases}}}\)

T=116x+14y+1zT=116x+14y+1z  ; x + y + z = 1

⇒T=x+y+z16x+x+y+z4y+x+y+zz⇒T=x+y+z16x+x+y+z4y+x+y+zz

=116+y16x+z16x+x4y+14+z4y+xz+yz+1=116+y16x+z16x+x4y+14+z4y+xz+yz+1

=(116+14+1)+(y16x+x4y)+(z16x+xz)+(z4y+yz)=(116+14+1)+(y16x+x4y)+(z16x+xz)+(z4y+yz)                    (1)

x;y;z>0⇒y16x;x4y;z16x;xz;z4y;yz>0x;y;z>0⇒y16x;x4y;z16x;xz;z4y;yz>0

áp dụng bđt cô si : 

y16x+x4y≥2√y16x⋅x4y=14y16x+x4y≥2y16x⋅x4y=14                             (2)

z16x+xz≥2√z16x⋅xz=12z16x+xz≥2z16x⋅xz=12                                 (3)

x4y+yz≥2√z4y⋅yz=1x4y+yz≥2z4y⋅yz=1                                        (4)

(1)(2)(3)(4) ⇒T≥116+14+1+14+12+1⇒T≥116+14+1+14+12+1

⇒T≥4916⇒T≥4916

dấu "=" xảy ra khi \hept⎧⎪ ⎪⎨⎪ ⎪⎩y16x=x4yz16x=xzz4y=yz⇔\hept⎧⎨⎩4y2=16x2z2=16x2z2=4y2\hept{y16x=x4yz16x=xzz4y=yz⇔\hept{4y2=16x2z2=16x2z2=4y2

⇔\hept⎧⎨⎩y=2xz=4xz=2y⇔\hept{y=2xz=4xz=2y có x+y+z = 1

=> x + 2x + 4x = 1

=> x = 1/7

xong tìm ra y = 2/7 và z = 4/7

T=116x+14y+1zT=116x+14y+1z  ; x + y + z = 1

⇒T=x+y+z16x+x+y+z4y+x+y+zz⇒T=x+y+z16x+x+y+z4y+x+y+zz

=116+y16x+z16x+x4y+14+z4y+xz+yz+1=116+y16x+z16x+x4y+14+z4y+xz+yz+1

=(116+14+1)+(y16x+x4y)+(z16x+xz)+(z4y+yz)=(116+14+1)+(y16x+x4y)+(z16x+xz)+(z4y+yz)                    (1)

x;y;z>0⇒y16x;x4y;z16x;xz;z4y;yz>0x;y;z>0⇒y16x;x4y;z16x;xz;z4y;yz>0

áp dụng bđt cô si : 

y16x+x4y≥2√y16x⋅x4y=14y16x+x4y≥2y16x⋅x4y=14                             (2)

z16x+xz≥2√z16x⋅xz=12z16x+xz≥2z16x⋅xz=12                                 (3)

x4y+yz≥2√z4y⋅yz=1x4y+yz≥2z4y⋅yz=1                                        (4)

(1)(2)(3)(4) ⇒T≥116+14+1+14+12+1⇒T≥116+14+1+14+12+1

⇒T≥4916⇒T≥4916

dấu "=" xảy ra khi \hept⎧⎪ ⎪⎨⎪ ⎪⎩y16x=x4yz16x=xzz4y=yz⇔\hept⎧⎨⎩4y2=16x2z2=16x2z2=4y2\hept{y16x=x4yz16x=xzz4y=yz⇔\hept{4y2=16x2z2=16x2z2=4y2

⇔\hept⎧⎨⎩y=2xz=4xz=2y⇔\hept{y=2xz=4xz=2y có x+y+z = 1

=> x + 2x + 4x = 1

=> x = 1/7

xong tìm ra y = 2/7 và z = 4/7

Lời giải:

Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}\right)(x+y+z)\geq \left(\sqrt{\frac{1}{16}}+\sqrt{\frac{1}{4}}+\sqrt{1}\right)^2\)

\(\Leftrightarrow P(x+y+z)\geq \frac{49}{16}\)

\(\Leftrightarrow P\geq \frac{49}{16}\) (do \(x+y+z=1\) )

Vậy \(P_{\min}=\frac{49}{16}\) tại \((x,y,z)=(\frac{1}{7}; \frac{2}{7}; \frac{4}{7})\)

x,y,z>0 nha

\(M=\dfrac{1}{16x^2}+\dfrac{1}{4y^2}+\dfrac{1}{16z^2}=\dfrac{1}{16}\left(\dfrac{1}{x^2}+\dfrac{2^2}{y^2}+\dfrac{4^2}{z^2}\right)\)

\(\Rightarrow M\ge\dfrac{1}{16}.\dfrac{\left(1+2+4\right)^2}{\left(x^2+y^2+z^2\right)}=\dfrac{49}{16}\)

\(\Rightarrow M_{min}=\dfrac{49}{16}\) khi \(\dfrac{1}{x^2}=\dfrac{2}{y^2}=\dfrac{4}{z^2}\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{1}{7}\\y^2=\dfrac{2}{7}\\z^2=\dfrac{4}{7}\end{matrix}\right.\)

\(1,\) Áp dụng BĐT: \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\text{ và }\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Dấu \("="\Leftrightarrow x=y\)

\(A=\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\\ A\ge\dfrac{1}{2}\left(1+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(1+\dfrac{4}{a+b}\right)^2+17=\dfrac{25}{2}+17=\dfrac{59}{2}\\ \text{Dấu }"="\Leftrightarrow\left\{{}\begin{matrix}a+\dfrac{1}{a}=b+\dfrac{1}{b}\\a+b=1\end{matrix}\right.\Leftrightarrow a=b=\dfrac{1}{2}\)

\(2,\text{Đặt }A=\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(\dfrac{xy^2z}{xz}+\dfrac{xyz^2}{xy}+\dfrac{x^2yz}{yz}\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(x^2+y^2+z^2\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+6\)

Áp dụng Cosi: \(\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}\ge2y^2\)

CMTT: \(\left\{{}\begin{matrix}\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}\ge2z^2\\\dfrac{x^2y^2}{z^2}+\dfrac{x^2z^2}{y^2}\ge2x^2\end{matrix}\right.\)

Cộng VTV \(\Leftrightarrow A^2\ge2\left(x^2+y^2+z^2\right)+6=12\\ \Leftrightarrow A\ge2\sqrt{3}\)

Dấu \("="\Leftrightarrow x=y=z=1\)

Svac-xơ nhé 

\(P=\frac{1}{16x}+\frac{4}{16y}+\frac{16}{16z}\ge\frac{\left(1+2+4\right)^2}{16\left(x+y+z\right)}=\frac{49}{16}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\frac{1}{16x}=\frac{2}{16y}=\frac{4}{16z}\)\(\Leftrightarrow\)\(\frac{1}{x}=\frac{2}{y}=\frac{4}{z}=\frac{1+2+4}{x+y+z}=7\)

Suy ra \(\hept{\begin{cases}x=\frac{1}{7}\\y=\frac{2}{7}\\z=\frac{4}{7}\end{cases}}\)

... 

\(P=\dfrac{\left(\dfrac{1}{4}\right)^2}{x}+\dfrac{\left(\dfrac{1}{2}\right)^2}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{4}+\dfrac{1}{2}+1\right)^2}{x+y+z}=\dfrac{49}{16}\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\dfrac{1}{7};\dfrac{2}{7};\dfrac{4}{7}\right)\)

x, y, z > 0 chứ bn ? Nếu đúng z thì inbox với mik, mik sẽ chỉ cho....

mình đánh nhầm bạn ạ. x y z >0 đấy