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hùi nãy mem nào k sai cho t T_T t buồn
\(VT\ge6\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-2\left(xy+yz+zx\right)+2.\frac{9}{4\left(x+y+z\right)}\)
\(=6\left(x+y+z\right)^2-2.\frac{\left(x+y+z\right)^2}{3}+\frac{9}{2\left(x+y+z\right)}=6.\left(\frac{3}{4}\right)^2-2.\frac{\left(\frac{3}{4}\right)^2}{3}+\frac{9}{2.\frac{3}{4}}\)
\(=\frac{27}{8}-\frac{3}{8}+6=9\)
\(\Rightarrow\)\(VT\ge9\) ( đpcm )
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{4}\)
Chúc bạn học tốt ~
\(VT=6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(=6\left(x+y+z\right)^2-2\left(xy+yz+xz\right)+2\frac{9}{2x+y+z+x+2y+z+x+y+2z}\)
\(\ge6\left(x+y+z\right)^2-2\frac{\left(x+y+z\right)^2}{3}+2\frac{9}{4\left(x+y+z\right)}\)
\(=\: 6\cdot\left(\frac{3}{4}\right)^2-2\cdot\frac{\left(\frac{3}{4}\right)^2}{3}+2\cdot\frac{9}{4\cdot\frac{3}{4}}=9\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT=\frac{2x^2+y^2+z^2}{4-yz}+\frac{2y^2+z^2+x^2}{4-xz}+\frac{2z^2+x^2+y^2}{4-xy}\)
\(\ge\frac{4x\sqrt{yz}}{4-yz}+\frac{4y\sqrt{xz}}{4-xz}+\frac{4z\sqrt{xy}}{4-xy}\)
Cần chứng minh \(\frac{4x\sqrt{yz}}{4-yz}+\frac{4y\sqrt{xz}}{4-xz}+\frac{4z\sqrt{xy}}{4-xy}\ge4xyz\)
\(\Leftrightarrow\frac{\sqrt{yz}}{yz\left(4-yz\right)}+\frac{\sqrt{xz}}{xz\left(4-xz\right)}+\frac{\sqrt{xy}}{xy\left(4-xy\right)}\ge1\)
Cauchy-Schwarz: \(\left(x+y+z\right)^2\ge\left(1+1+1\right)\left(xy+yz+xz\right)\ge\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)^2\)
\(\Leftrightarrow3\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\)
Đặt \(\left(\sqrt{xy};\sqrt{yz};\sqrt{xz}\right)\rightarrow\left(a;b;c\right)\)\(\Rightarrow\hept{\begin{cases}a,b,c>0\\a+b+c\le3\end{cases}}\)
\(\Leftrightarrow\frac{a}{a^2\left(4-a^2\right)}+\frac{b}{b^2\left(4-b^2\right)}+\frac{c}{c\left(4-c^2\right)}\ge1\left(\odot\right)\)
Ta có BĐT phụ: \(\dfrac{a}{a^2\left(4-a^2\right)}\le-\dfrac{1}{9}a+\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{\left(a-1\right)^2\left(a^2-2a-9\right)}{9a\left(a-2\right)\left(a+2\right)}\le0\forall0< a\le1\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế
\(VT_{\left(\odot\right)}\ge\dfrac{-\left(a+b+c\right)}{9}+\dfrac{4}{9}\cdot3\ge\dfrac{-3}{9}+\dfrac{12}{9}=1=VP_{\left(\odot\right)}\)
Dấu "=" <=> x=y=z=1
Xem lại cái đề đi Tuyển. Hình như giá trị nhỏ nhất của cái biểu thức dưới còn lớn hơn là 1 thì làm sao bài đó có giá trị x, y, z thỏa được mà bảo tính A.
Ta có : 2P = \(\frac{\sqrt{4x^2-4xy+4y^2}}{x+y+2z}+\frac{\sqrt{4y^2-4yz+4z^2}}{y+z+2x}+\frac{\sqrt{4z^2-4zx+4x^2}}{z+x+2y}\)
\(=\frac{\sqrt{\left(2x-y\right)^2+\left(\sqrt{3}y\right)^2}}{x+y+2z}+\frac{\sqrt{\left(2y-z\right)^2+\left(\sqrt{3}z\right)^2}}{y+z+2x}+\frac{\sqrt{\left(2z-x\right)^2+\left(\sqrt{3}x\right)^2}}{z+x+2y}\)
Lại có \(\frac{\sqrt{\left[\left(2x-y\right)^2+\left(\sqrt{3}y\right)^2\right]\left[\left(1^2+\left(\sqrt{3}\right)^2\right)\right]}}{x+y+2z}\ge\frac{\left[\left(2x-y\right).1+3y\right]}{x+y+2z}=\frac{2\left(x+y\right)}{x+y+2z}\)
=> \(\sqrt{\frac{\left(2x-y\right)^2+\left(\sqrt{3}y\right)^2}{x+y+2z}}\ge\frac{x+y}{x+y+2z}\)(BĐT Bunyakovsky)
Tương tự ta đươc \(2P\ge\frac{x+y}{x+y+2z}+\frac{y+z}{2x+y+z}+\frac{z+x}{2y+z+x}\)
Đặt x + y = a ; y + z = b ; x + z = c
Khi đó \(2P\ge\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)
\(\ge\left(a+b+c\right).\frac{9}{2\left(a+b+c\right)}-3\ge\frac{9}{2}-3=\frac{3}{2}\)
=> \(P\ge\frac{3}{4}\)
Dấu "=" xảy ra <=> x = y = z
bài 8 : bỏ dấu hoặc rồi tính
a;( 17 - 299) + ( 17 - 25 + 299)
Thấy cái đề mà thấy khiếp ...
Ta có : \(x^2-xy+y^2=\frac{3}{4}\left(x^2-2xy+y^2\right)+\frac{1}{4}\left(x^2+2xy+y^2\right)\)
\(=\frac{3}{4}\left(x-y\right)^2+\frac{1}{4}\left(x+y\right)^2\ge\frac{1}{4}\left(x+y\right)^2\)
\(\Rightarrow\sqrt{x^2-xy+y^2}\ge\frac{x+y}{2}\)
Tương tự \(\sqrt{y^2-yz+z^2}\ge\frac{y+z}{2}\)
\(\sqrt{z^2-zx+x^2}\ge\frac{x+z}{2}\)
Do đó : \(2S\ge\frac{x+y}{x+y+2z}+\frac{y+z}{y+z+2x}+\frac{x+z}{x+z+2y}\)
\(\Rightarrow2S+3\ge\left(1+\frac{x+y}{x+y+2z}\right)+\left(1+\frac{y+z}{y+z+2x}\right)+\left(1+\frac{x+z}{x+z+2y}\right)\)
\(=2\left(x+y+z\right)\left(\frac{1}{x+y+2z}+\frac{1}{y+z+2x}+\frac{1}{x+z+2y}\right)\)
\(\ge2\left(x+y+z\right).\frac{9}{4\left(x+y+z\right)}\)\(=\frac{9}{2}\)
(Áp dụng bđt Cô-si dạng engel cho 3 số)
\(\Rightarrow2S+3\ge\frac{9}{2}\)
\(\Rightarrow S\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)
Vậy ..............
\(6\left(x^2+y^2+z^2\right)+10\left(xy+yz+zx\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(=5\left(x+y+z\right)^2+\left(x^2+y^2+z^2\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(\ge5.\left(\frac{3}{4}\right)^2+\frac{\left(x+y+z\right)^2}{3}+\frac{2.9}{4\left(x+y+z\right)}\)
\(=5.\left(\frac{3}{4}\right)^2+\frac{\left(\frac{3}{4}\right)^2}{3}+\frac{2.9}{\frac{4.3}{4}}=9\)