Question

[monsiaur kite]

với x+y+z=1và x;y;z>0,tìm Min của biểu thức sau:

\(S=\left(\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}\right)\)

Answer 1

[✎﹏ミ★꧁༺вєѕт↭ℓαυяιєℓ↭νи༻꧂★ミ.༻(Trưở...]

\(S=\left(\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}\right)=\left(\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}\right).\left(x+y+z\right)\)   (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 \(\Leftrightarrow\hept{\begin{cases}4x=2y=z\\x+y+z=1\\x;y;z>0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{7}\\y=\frac{2}{7}\\z=\frac{4}{7}\end{cases}}}\)

Answer 2

[Hotory Yaruyuki]

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