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Phải sửa xy/x^2+y^2 thành x^2+y^2/xy hoặc cái gì khác
Vì xy/x^2+y^2 chỉ có GTLN chứ ko có GTNN đâu
Mk nói có gì sai thì thông cảm nha !
Ta có \(ax^3=by^3=cz^3\Leftrightarrow\dfrac{ax^2}{\dfrac{1}{x}}=\dfrac{by^2}{\dfrac{1}{y}}=\dfrac{cz^2}{\dfrac{1}{z}}=\dfrac{ax^2+by^2+cz^2}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=ax^2+by^2+cz^2\Leftrightarrow\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{ax^3}=\sqrt[3]{by^3}=\sqrt[3]{cz^3}=\dfrac{\sqrt[3]{a}}{\dfrac{1}{x}}+\dfrac{\sqrt[3]{b}}{\dfrac{1}{y}}+\dfrac{\sqrt[3]{c}}{\dfrac{1}{z}}=\dfrac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)Vậy \(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)
Ta co : (x+y)2≤2(x2+y2)
=> x+y≤\(\sqrt{2\left(x^2+y^2\right)}\)
=> \(\dfrac{z^2}{x+y}\ge\dfrac{z^2}{\sqrt{2\left(x^2+y^2\right)}}\)
Tuong tu: \(\dfrac{x^2}{y+z}\ge\dfrac{x^2}{\sqrt{2\left(y^2+z^2\right)}}\)
\(\dfrac{y^2}{x+z}\ge\dfrac{y^2}{\sqrt{2\left(x+z\right)}}\)
VT≥\(\dfrac{x^2}{\sqrt{2\left(y^2+z^2\right)}}+\dfrac{y^2}{\sqrt{2\left(x^2+z^2\right)}}+\dfrac{z^2}{\sqrt{2\left(x^2+y^2\right)}}\)
Dat : \(\sqrt{y^2+z^2}=a\)
\(\sqrt{x^2+z^2}=b\)
\(\sqrt{x^2+y^2}=c\)
=> a+b+c=2015 , a2=y2+z2 , b2=x2+z2 , c2=x2+y2
=> VT≥ \(\dfrac{b^2+c^2-a^2}{2\sqrt{2}.a}+\dfrac{a^2+c^2-b^2}{2\sqrt{2}.b}+\dfrac{a^2+b^2-c^2}{2\sqrt{2}c}\)
≥ \(\dfrac{1}{2\sqrt{2}}\left[\dfrac{\left(b+c\right)^2}{2a}+\dfrac{\left(a+b\right)^2}{2c}+\dfrac{\left(a+c\right)^2}{2b}-2015\right]\)
≥\(\dfrac{1}{2\sqrt{2}}\left[2\left(a+b+c\right)-2015\right]\)
= \(\dfrac{2015}{2\sqrt{2}}\)
\(\text{a) }\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}\\ =\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)-2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)}\\ =\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\cdot\dfrac{x+y+z}{xyz}}\\ =\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\)
\(\text{b) }\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\\ =1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2017}-\dfrac{1}{2018}\\ =2016+\dfrac{1}{2}-\dfrac{1}{2018}\\ =\dfrac{2034698}{1009}\)
Ta có: \(\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy\)
\(=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)
\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)\(\ge4+2+1=7\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
Vậy \(\left(\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy\right)_{Min}=7\Leftrightarrow x=y=\frac{1}{2}\)
à nhầm, bạn pham trung thanh làm đúng rồi đấy mọi người ủng hộ bạn ấy nha
theo bđt cauchy schwars dạng engel ta có
\(T=\dfrac{x^2}{y+x}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\)
Dấu '=' xảy ra khi x=y=z
pt \(\Leftrightarrow\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}=2015\)
\(\Leftrightarrow3\sqrt{2}x=2015\)
\(\Leftrightarrow x=\dfrac{2015}{3\sqrt{2}}\)
vậy \(T_{min}=\dfrac{2015}{\sqrt{2}}\) khi \(x=y=z=\dfrac{2015}{3\sqrt{2}}\)
ko chắc đúng nha bạn :))