1 thách dám tích

\(A=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\ge\frac{4}{\left(x+y\right)^2}+\frac{1}{2xy}\\ =\frac{1}{4}+\frac{1}{2xy}\ge\frac{1}{4}+\frac{1}{8}=\frac{3}{8}\)

Dấu = xảy ra khi x=y=2

By Titu's Lemma we easy have:

\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)

\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{17}{4}\)

Mk xin b2 nha!

\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)

\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)

\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)

\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

\(\frac{1}{x^2}+\frac{1}{9y^2}\ge2\sqrt{\frac{1}{x^2}.\frac{1}{9y^2}}=\frac{2}{3xy}=\frac{2}{3}\)

Dấu \(=\)xảy ra khi \(\hept{\begin{cases}\frac{1}{x^2}=\frac{1}{9y^2}\\xy=1\end{cases}}\Rightarrow\hept{\begin{cases}x=\sqrt{3}\\y=\frac{1}{\sqrt{3}}\end{cases}}\).

Bài 3:

Áp dụng BĐT Bunhiacopxky như các bài trên ta có:

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$

Mà $0< x+y+z\leq 6$ nên $\frac{9}{x+y+z}\geq \frac{9}{6}=\frac{3}{2}$

Do đó $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{3}{2}$ (đpcm)

Dấu "=" xảy ra khi $x=y=z=2$

Bài 4:

Áp dụng BĐT Cô-si cho các số dương ta có:

$a^4+b^4+c^4+d^4\geq 4\sqrt[4]{a^4b^4c^4d^4}=4abcd$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=d>0$

Bài 2:

Hoàn toàn tương tự bài 1

$S(a+b+c)\geq (1+1+1)^2$ theo BĐT Bunhiacopxky

$\Leftrightarrow S.3\geq 9\Rightarrow S\geq 3$

Vậy GTNN của $S$ là $3$ khi $a=b=c=1$

\(xy+yz+zx=4xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\)

Ta có \(M=\frac{1}{4\left(x+y\right)}+\frac{1}{4\left(y+z\right)}+\frac{1}{4\left(z+x\right)}\)

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

                                                                                     \(=\frac{1}{8}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{8}.4=\frac{1}{2}\)

Dấu "=" tại x = y = z = ³/₄

\(P=\frac{x\left(x+5\right)+y\left(y+5\right)+2\left(xy-3\right)}{x\left(x+6\right)+y\left(y+6\right)+2xy}\)

\(=\frac{x^2+5x+y^2+5y+2xy-6}{x^2+6x+y^2+6y+2xy}\)

\(=\frac{\left(x+y\right)^2+5\left(x+y\right)-6}{\left(x+y\right)^2+6\left(x+y\right)}\)

\(=\frac{\left(x+y\right)\left(x+y+5\right)-6}{\left(x+y\right)\left(x+y+6\right)}\)

\(=\frac{2005\times\left(2005+5\right)-6}{2005\times\left(2005+6\right)}\)

\(=\frac{2005\times2010-6}{2005\times2011}\)

\(=\frac{2004}{2005}\)