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a) \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)
\(\Leftrightarrow2x^2+2y^2\ge\left(x+y\right)^2\Leftrightarrow x^2+y^2\ge2xy\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\left(đúng\right)\)
b) \(x^3+y^3\ge\dfrac{\left(x+y\right)^3}{4}\)
\(\Leftrightarrow4x^3+4y^3\ge\left(x+y\right)^3\Leftrightarrow3x^3+3y^3\ge3x^2y+3xy^2\)
\(\Leftrightarrow3x^2\left(x-y\right)-3y^2\left(x-y\right)\ge0\)
\(\Leftrightarrow3\left(x-y\right)\left(x^2-y^2\right)\ge0\Leftrightarrow3\left(x-y\right)^2\left(x+y\right)\ge0\left(đúng\right)\)
a: Ta có: \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)
\(\Leftrightarrow2x^2+2y^2-x^2-2xy-y^2\ge0\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\)(luôn đúng)
a)Áp dụng BĐT AM-GM ta có:
\(\left(\sqrt{x}+\sqrt{y}\right)^2=x+y+2\sqrt{xy}\)
\(\ge2\sqrt{\left(x+y\right)\cdot2\sqrt{xy}}=VP\)
Xảy ra khi \(x=y\)
b)\(BDT\Leftrightarrow x+y+z+t\ge4\sqrt[4]{xyzt}\)
Đúng với AM-GM 4 số
Xảy ra khi \(x=y=z=t\)
\(x^4+y^4\ge\frac{\left(x^2+y^2\right)^2}{2}\ge\frac{\left[\frac{\left(x+y\right)^2}{2}\right]^2}{2}=\frac{\left(x+y\right)^4}{8}\)(bđt Cauchy - Schwarz)
Chứng minh bằng biến đổi tương đương:
\(x^8+y^8\ge x^2y^2\left(x^4+y^4\right)\)
\(\Leftrightarrow x^8-x^6y^2+y^8-x^2y^6\ge0\)
\(\Leftrightarrow x^6\left(x^2-y^2\right)-y^6\left(x^2-y^2\right)\ge0\)
\(\Leftrightarrow\left(x^6-y^6\right)\left(x^2-y^2\right)\ge0\)
\(\Leftrightarrow\left[\left(x^2\right)^3-\left(y^2\right)^3\right]\left(x^2-y^2\right)\ge0\)
\(\Leftrightarrow\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)\left(x^2-y^2\right)\ge0\)
\(\Leftrightarrow\left(x^2-y^2\right)^2\left(x^4+x^2y^2+y^4\right)\ge0\) (luôn đúng với mọi x;y)
Vậy BĐT đã cho được chứng minh.
Ta có \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\).
Suy ra \(x^4+y^4\ge\dfrac{\left(x^2+y^2\right)^2}{2}\)\(\ge\dfrac{\left[\dfrac{\left(x+y\right)^2}{2}\right]^2}{2}=\dfrac{\left(x+y\right)^4}{8}\). (đpcm).
Lời giải:
Xét hiệu \((x+y)\left(\frac{1}{x}+\frac{1}{y}\right)-4=\left(1+\frac{x}{y}+\frac{y}{x}+1\right)-4\)
\(=\frac{x}{y}+\frac{y}{x}-2=\frac{x^2+y^2}{xy}-2=\frac{x^2+y^2-2xy}{xy}=\frac{(x-y)^2}{xy}\geq 0, \forall x,y>0\)
Do đó \((x+y)\left(\frac{1}{x}+\frac{1}{y}\right)\geq 4\) (đpcm)
Dấu "=" xảy ra khi \((x-y)^2=0\Leftrightarrow x=y\)
x,y khác 0 đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}81a+105b=8\\54a+42b=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}9.9.2a+105.2.b=8.2\\9.6.3a+42.3b=4.3\end{matrix}\right.\)
\(\Leftrightarrow\left(105.2-42.3\right)b=8.2-4.3=4\left(4-3\right)=4\)
\(\Leftrightarrow\left(105-21.3\right)b=2\)
\(\Leftrightarrow3\left(35-21\right)b=2\Rightarrow b=\dfrac{2}{3.14}=\dfrac{1}{3.7}=\dfrac{1}{21}\)
\(54a+42.\dfrac{1}{21}=4\Leftrightarrow54a+2=4\)
\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\a=\dfrac{1}{27}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=27\\y=21\end{matrix}\right.\)
