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Từ \(\dfrac{a}{1+a}+\dfrac{2b}{2+b}+\dfrac{3c}{3+c}\le\dfrac{6}{7}\)
\(\Leftrightarrow1-\dfrac{a}{1+a}+2-\dfrac{2b}{2+b}+3-\dfrac{3c}{3+c}\ge6-\dfrac{6}{7}\)
\(\Leftrightarrow\dfrac{1}{a+1}+\dfrac{4}{b+2}+\dfrac{9}{c+3}\ge\dfrac{36}{7}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{1}{a+1}+\dfrac{4}{b+2}+\dfrac{9}{c+3}\)
\(\ge\dfrac{\left(1+2+3\right)^2}{a+b+c+6}=\dfrac{36}{7}=VP\)
Xảy ra khi \(a=\dfrac{1}{6};b=\dfrac{1}{3};c=\dfrac{1}{2}\)
2) \(\dfrac{1}{x}+\dfrac{25}{y}+\dfrac{64}{z}=\dfrac{4}{4x}+\dfrac{225}{9y}+\dfrac{1024}{16z}\ge\dfrac{\left(2+15+32\right)^2}{4x+9y+6z}=49\)
\(VT=\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)^2\)
\(VT\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{25}{2}\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)
Do \(x;y>0\) ta biến đổi tương đương:
\(\dfrac{x+y}{2}\ge\dfrac{2}{\dfrac{1}{x}+\dfrac{1}{y}}\Leftrightarrow\dfrac{x+y}{2}\ge\dfrac{2}{\dfrac{x+y}{xy}}\)
\(\Leftrightarrow\dfrac{x+y}{2}\ge\dfrac{2xy}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow x^2+2xy+y^2\ge4xy\Leftrightarrow x^2-2xy+y^2\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
Vậy BĐT được chứng minh, dấu "=" xảy ra khi \(x=y\)
Lời giải:
Ta có: \(xy+yz+xz=3xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
Mà theo BĐT Cauchy-Schwarz: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}\)
Do đó: \(3\geq \frac{9}{x+y+z}\Rightarrow x+y+z\geq 3\)
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Ta có: \(\text{VT}=x-\frac{xz}{x^2+z}+y-\frac{xy}{y^2+x}+z-\frac{yz}{z^2+y}\)
\(=(x+y+z)-\left(\frac{xy}{y^2+x}+\frac{yz}{z^2+y}+\frac{xz}{x^2+z}\right)\)
\(\geq x+y+z-\frac{1}{2}\left(\frac{xy}{\sqrt{xy^2}}+\frac{yz}{\sqrt{z^2y}}+\frac{xz}{\sqrt{x^2z}}\right)\) (AM-GM)
\(=x+y+z-\frac{1}{2}(\sqrt{x}+\sqrt{y}+\sqrt{z})\)
Tiếp tục AM-GM: \(\sqrt{x}+\sqrt{y}+\sqrt{z}\leq \frac{x+1}{2}+\frac{y+1}{2}+\frac{z+1}{2}=\frac{x+y+z+3}{2}\)
Suy ra:
\(\text{VT}\geq x+y+z-\frac{1}{2}.\frac{x+y+z+3}{2}=\frac{3}{4}(x+y+z)-\frac{3}{4}\)
\(\geq \frac{9}{4}-\frac{3}{4}=\frac{3}{2}=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Ta có đpcm
Dấu bằng xảy ra khi $x=y=z=1$
\(VT=\dfrac{x\sqrt{y}+y\sqrt{x}}{x+y}-\dfrac{x+y}{2}\le\dfrac{\sqrt{2xy\left(x+y\right)}}{x+y}-\dfrac{x+y}{2}\)
\(\le\dfrac{\left(x+y\right)\sqrt{\dfrac{x+y}{2}}}{x+y}-\dfrac{x+y}{2}\). Cần cm \(\sqrt{\dfrac{x+y}{2}}-\dfrac{x+y}{2}\le\dfrac{1}{4}\)
Đặt \(x+y=t>0\) thì:
\(\sqrt{\dfrac{t}{2}}-\dfrac{t}{2}\le\dfrac{1}{4}\Leftrightarrow-\dfrac{1}{4}\left(\sqrt{2t}-1\right)^2\le0\) *Đúng*
Áp dụng bất đẳng thức Côsi cho các số dương $x, y, z$, ta được:$x^{3}+y^{2} \geqslant 2 \sqrt{x^{3} \cdot y^{2}}=2 x y \cdot \sqrt{x}$$y^{3}+z^{2} \geqslant 2 \sqrt{y^{3} \cdot z^{2}}=2 y z \cdot \sqrt{y}$$z^{3}+x^{2} \geqslant 2 \sqrt{z^{3} \cdot x^{2}}=2 z x \cdot \sqrt{z}$Khi đó BĐT đã cho trở thành:$\dfrac{2 \sqrt{x}}{x^{3}+y^{2}}+\dfrac{2 \sqrt{y}}{y^{3}+z^{2}}+\dfrac{2 \sqrt{z}}{z^{3}+x^{2}} \leqslant \dfrac{2 \sqrt{x}}{2 x y \sqrt{x}}+\dfrac{2 \sqrt{y}}{2 y z \sqrt{y}}+\dfrac{2 \sqrt{z}}{2 z x \sqrt{z}}=\dfrac{1}{x y}+\dfrac{1}{y z}+\dfrac{1}{z x} (1)$Mặt khác ta có:$\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}} \geqslant \dfrac{2}{x y} \Rightarrow \dfrac{1}{x y} \leqslant \dfrac{1}{2}\left(\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}\right)$
CMTT: $\dfrac{1}{y z} \leq \dfrac{1}{2}\left(\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}\right) ; \dfrac{1}{z x} \leqslant \dfrac{1}{2}\left(\dfrac{1}{z^{2}}+\dfrac{1}{x^{2}}\right)$Suy ra: $\dfrac{1}{x y}+\dfrac{1}{y z}+\dfrac{1}{z x} \leqslant \dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}(2)$Từ $(1)$ và $(2)$ ta được: $\dfrac{2 \sqrt{x}}{x^{3}+y^{2}}+\dfrac{2 \sqrt{y}}{y^{3}+z^{2}}+\dfrac{2 \sqrt{z}}{z^{3}+x^{2}} \leqslant \dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}$Dấu " $="$ xảy ra $\Leftrightarrow x=y=z=1$
Áp dụng bất đẳng thức Côsi cho các số dương $x, y, z$, ta được:
$x^{3}+y^{2} \geqslant 2 \sqrt{x^{3} \cdot y^{2}}=2 x y \cdot \sqrt{x}$
$y^{3}+z^{2} \geqslant 2 \sqrt{y^{3} \cdot z^{2}}=2 y z \cdot \sqrt{y}$
$z^{3}+x^{2} \geqslant 2 \sqrt{z^{3} \cdot x^{2}}=2 z x \cdot \sqrt{z}$
Khi đó BĐT đã cho trở thành:
$\dfrac{2 \sqrt{x}}{x^{3}+y^{2}}+\dfrac{2 \sqrt{y}}{y^{3}+z^{2}}+\dfrac{2 \sqrt{z}}{z^{3}+x^{2}} \leqslant \dfrac{2 \sqrt{x}}{2 x y \sqrt{x}}+\dfrac{2 \sqrt{y}}{2 y z \sqrt{y}}+\dfrac{2 \sqrt{z}}{2 z x \sqrt{z}}=\dfrac{1}{x y}+\dfrac{1}{y z}+\dfrac{1}{z x} (1)$
Mặt khác ta có:
$\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}} \geqslant \dfrac{2}{x y} \Rightarrow \dfrac{1}{x y} \leqslant \dfrac{1}{2}\left(\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}\right)$
CMTT: $\dfrac{1}{y z} \leq \dfrac{1}{2}\left(\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}\right) ; \dfrac{1}{z x} \leqslant \dfrac{1}{2}\left(\dfrac{1}{z^{2}}+\dfrac{1}{x^{2}}\right)$
Suy ra: $\dfrac{1}{x y}+\dfrac{1}{y z}+\dfrac{1}{z x} \leqslant \dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}(2)$
Từ $(1)$ và $(2)$ ta được: $\dfrac{2 \sqrt{x}}{x^{3}+y^{2}}+\dfrac{2 \sqrt{y}}{y^{3}+z^{2}}+\dfrac{2 \sqrt{z}}{z^{3}+x^{2}} \leqslant \dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}$
Dấu " $="$ xảy ra $\Leftrightarrow x=y=z=1$
1) \(\dfrac{x}{3}=\dfrac{y}{4}=t\Leftrightarrow\left\{{}\begin{matrix}x=3t\\y=4t\end{matrix}\right.\)
ta có \(x.y^2=324\Leftrightarrow3t.\left(4t\right)^2=324\)
\(\Leftrightarrow t^3=\dfrac{27}{4}\)
\(\Leftrightarrow t=\dfrac{3}{\sqrt[3]{4}}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3.\dfrac{3}{\sqrt[3]{4}}=\dfrac{9}{\sqrt[3]{4}}\\y=4.\dfrac{3}{\sqrt[3]{4}}=\dfrac{12}{\sqrt[3]{4}}\end{matrix}\right.\)
2) \(2^{x+1}.3^y=2^{2x}.3^x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
3) \(\dfrac{a}{b}=\dfrac{c}{d}\)
áp dụng dãy tỉ số = nhau ta có
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a-c}{b-d}\)
\(\Leftrightarrow\dfrac{a^4}{b^4}=\dfrac{c^4}{d^4}=\left(\dfrac{a-c}{b-d}\right)^4\left(1\right)\)
mà \(\dfrac{a^4}{b^4}=\dfrac{c^4}{d^4}=\dfrac{a^4+c^4}{b^4+c^4}\left(2\right)\)
từ (1)(2) suy ra đpcm
4) \(B=\dfrac{27^{15}.5^3.8^4}{25^2.81^{11}.2^{11}}=\dfrac{\left(3^3\right)^{15}.5^3.\left(2^3\right)^4}{\left(5^2\right)^2.\left(3^4\right)^{11}.2^{11}}=\dfrac{3^{45}.5^3.2^{12}}{5^4.3^{44}.2^{11}}=\dfrac{3.2}{5}=\dfrac{6}{5}\)
Ta có:
\(1=\dfrac{2^2}{x}+\dfrac{5^2}{y}\ge\dfrac{\left(2+5\right)^2}{x+y}=\dfrac{49}{x+y}\)
\(\Rightarrow x+y\ge49\) (đpcm)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(14;35\right)\)