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Ta có : \(8^x+8^x+8^2\ge3\sqrt[3]{8^x.8^x.8^2}=12.4^x\)
\(8^y+8^y+8^2\ge3\sqrt[3]{8^y.8^y.8^2}=12.4^y\)
\(8^z+8^z+8^2\ge3\sqrt[3]{8^z.8^z.8^2}=12.4^z\)
\(8^x+8^y+8^z\ge3\sqrt[3]{8^x.8^y.8^z}=3\sqrt[3]{8^6}=192\)
Cộng các vế , ta được :
\(3\left(8^x+8^y+8^z+64\right)\ge3\left(4^{x+1}+4^{y+1}+4^{z+1}+64\right)\)
hay \(8^x+8^y+8^z\ge4^{x+1}+4^{y+1}+4^{z+1}\)
Dự đoán dấu bằng xảy ra khi \(x=y=z=2\), áp dụng BĐT AM-GM ta có:
\(8^x+8^x+64\ge3\sqrt[3]{8^x\cdot8^x\cdot64}=12\cdot4^x\)
\(8^y+8^y+64\ge3\sqrt[3]{8^y\cdot8^y\cdot64}=12\cdot4^y\)
\(8^z+8^z+64\ge3\sqrt[3]{8^z\cdot8^z\cdot64}=12\cdot4^z\)
Suy ra \(2\left(8^x+8^y+8^z\right)+3\cdot64\ge12\left(4^x+4^y+4^z\right)\left(1\right)\)
Theo giả thiết ta có:
\(8^x+8^y+8^z\ge3\sqrt[3]{8^{x+y+z}}=3\sqrt[3]{8^6}=3\cdot64\left(2\right)\)
Cộng (1) với (2) theo vế ta có:
\(3\left(8^x+8^y+8^z\right)\ge12\left(4^x+4^y+4^z\right)=4^{x+1}+4^{y+1}+4^{z+1}\)
a/
-Cauchy-Schwar
\(P=\sum\frac{a^4}{a\sqrt{b^2+3}}\ge\frac{\left(\sum a^2\right)^2}{\sum a\sqrt{b^2+3}}\)
Côsi: \(\sum a\sqrt{b^2+3}=\frac{1}{2}\sum2a.\sqrt{b^2+3}\le\frac{1}{2}.\sum\frac{\left(2a\right)^2+b^2+3}{2}=\frac{1}{4}.\left[5\left(a^2+b^2+c^2\right)+3.3\right]=6\)
\(\Rightarrow P\ge\frac{3^2}{6}=\frac{3}{2}\)
Đẳng thức xảy ra khi a = b = c = 1.
b/
Côsi: \(8^x+8^x+64\ge3\sqrt[3]{8^x.8^x.64}=12.4^x\Rightarrow8^x\ge6.4^x-32\)
\(\Rightarrow8^x+8^y+8^z\ge6\left(4^x+4^y+4^z\right)-96\)
\(4^x+4^y+4^z\ge3\sqrt[3]{4^{x+y+z}}=3\sqrt[3]{4^6}=48\)
\(\Rightarrow-2\left(4^x+4^y+4^z\right)\le-96\)
\(\Rightarrow8^x+8^y+8^z\ge6\left(4^x+4^y+4^z\right)-2\left(4^x+4^y+4^z\right)=4^{x+1}+4^{y+1}+4^{z+1}\)
nx \(4\left(1-x\right)\left(1-y\right)\left(1-z\right)=4\left(y+z\right)\left(1-y\right)\left(1-z\right)\)
ap dung bdt \(\left(a+b\right)^2\ge4ab\) ta co \(4\left(y+z\right)\left(1-z\right)\left(1-y\right)\le\left(y+z+1-z\right)^2\left(1-y\right)=\left(y+1\right)^2\left(1-y\right)\) \(=\left(y+1\right)\left(y+1\right)\left(1-y\right)=\left(y+1\right)\left(1-y^2\right)\le y+1\) =\(y+x+y+z=x+2y+z\left(dpcm\right)\)
Bài 1:
Theo BĐT AM-GM có :$(x+y+1)(x^2+y^2)+\dfrac{4}{x+y}\geq (x+y+1).2xy+\dfrac{4}{x+y}=2(x+y+1)+\dfrac{4}{x+y}=(x+y)+(x+y)+\dfrac{4}{x+y}+2\geq 2\sqrt{xy}+2\sqrt{(x+y).\dfrac{4}{x+y}}+2=2+4+2=8$(đpcm)
Dấu \(=\) xảy ra khi \(x=y, xy=1\) và \(x+y=2\) hay \(x=y=1\)
Bài 1:
Áp dụng BĐT Cô-si cho các số dương:
\(x^2+y^2\geq 2xy=2\Rightarrow (x+y+1)(x^2+y^2)+\frac{4}{x+y}\geq 2(x+y+1)+\frac{4}{x+y}(1)\)
Tiếp tục áp dụng BĐT Cô-si:
\(2(x+y+1)+\frac{4}{x+y}=(x+y+2)+[(x+y)+\frac{4}{x+y}]\)
\(\geq (2\sqrt{xy}+2)+2\sqrt{(x+y).\frac{4}{x+y}}=(2+2)+4=8(2)\)
Từ \((1);(2)\Rightarrow (x+y+1)(x^2+y^2)+\frac{4}{x+y}\geq 8\) (đpcm)
Dấu "=" xảy ra khi $x=y=1$
BĐT đã cho <=> 1 + y \(\ge\) 4.(1 - x).(1 - y).(1 - z)
Áp dụng BĐT : 4ab \(\le\) (a + b)2 ta có: 4.(1 - x)(1 - z) \(\le\) (1 - x + 1 - z)2 = (1 + y)2
=> 4.(1 - x)(1 - y)(1 - z) \(\le\) (1 + y)2.(1 - y) = (1 + y).(1 -y2) \(\le\) (1 + y) .1 = 1+ y => đpcm
Dấu "=" xảy ra khi 1 - y2 = 1 và x = z => y = 0 ; x = z = 1/2
Áp dụng công thức \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\left(x,y>0\right)\)
Ta có \(\frac{1}{2x+y+z}\le\frac{1}{4}\left(\frac{1}{2x}+\frac{1}{y+z}\right)\)
\(\frac{1}{y+z}\le\frac{1}{4y}+\frac{1}{4z}\)
=> \(\frac{1}{2x+y+z}\le\frac{1}{4}\left(\frac{1}{2x}+\frac{1}{4y}+\frac{1}{4z}\right)\left(1\right)\)
Tương tự \(\hept{\begin{cases}\frac{1}{x+2y+z}\le\frac{1}{4}\left(\frac{1}{4x}+\frac{1}{2y}+\frac{1}{4z}\right)\left(2\right)\\\frac{1}{x+y+2z}\le\frac{1}{4}\left(\frac{1}{4x}+\frac{1}{4y}+\frac{1}{2z}\right)\left(3\right)\end{cases}}\)
(1)(2)(3) => \(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
=> \(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)
Dấu "=" xảy ra <=> \(x=y=z=\frac{3}{4}\)