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\(1,\)
Áp dụng BĐT Bunhiacopski:
\(A^2=\left(\sqrt{3-x}+\sqrt{x+7}\right)^2\le\left(1^2+1^2\right)\left(3-x+x+7\right)=2\cdot10=20\)
Dấu \("="\Leftrightarrow3-x=x+7\Leftrightarrow x=-2\)
\(A^2=3-x+x+7+2\sqrt{\left(3-x\right)\left(x+7\right)}\\ A^2=10+2\sqrt{\left(3-x\right)\left(x+7\right)}\ge10\)
Dấu \("="\Leftrightarrow\left(3-x\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)
Ta luôn có :
\(\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2\ge0\forall a,b\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}}\)
\(\Leftrightarrow2\left(\frac{1}{a}+\frac{1}{b}\right)\ge\frac{2}{\sqrt{ab}}+\frac{1}{a}+\frac{1}{b}\)
\(\Leftrightarrow\frac{2\left(a+b\right)}{ab}\ge\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)^2\)
\(\Leftrightarrow\sqrt{\frac{2\left(a+b\right)}{ab}}\ge\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế :
\(\sqrt{2}\left(\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{a+c}{ac}}\right)\)
\(\ge2\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\)
\(\Leftrightarrow\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{a+c}{ac}}\ge\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Chúc bạn học tốt !!!
Đặt \(\frac{1}{\sqrt{a}}=x,\frac{1}{\sqrt{b}}=y,\frac{1}{\sqrt{c}}\)=z
Thay vào ta có:\(\sqrt{2}\)(x+y+x)\(\le\)\(\sqrt{\left(x^2+y^2\right)}+\sqrt{x^2+z^2}+\sqrt{\left(y^2+z^2\right)}\)
Ta có bất đẳng thức sau A: (m2+n2)(p2+q2)\(\ge\)(mp+nq)2 dễ dàng chứng mình bằng cách khai triển
áp dụng bdt A với m=x,n=z,p=\(\sqrt{2}\).q=\(\sqrt{2}\) ta được
\(\sqrt{\frac{\left(x^2+z^2\right)\left(\sqrt{2}^2+\sqrt{2}^2\right)}{4}}\ge\sqrt{\left(x\sqrt{2}+z\sqrt{2}\right)^2}\)/2=\(\frac{\sqrt{2}\left(x+y\right)}{2}\)
Tương tự với cái phần tử còn lại ta được điều cần cm
Theo giả thiết, ta có: \(ab+bc+ca+abc=4\)
\(\Leftrightarrow abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8\)\(=12+\left(ab+bc+ca\right)+4\left(a+b+c\right)\)
\(\Leftrightarrow\left(a+2\right)\left(b+2\right)\left(c+2\right)\)\(=\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)\)
\(\Leftrightarrow\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1\)
\(\Rightarrow a+b+c+6=12\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)-6+a+b+c\)
\(=\left(\frac{12}{a+2}+a-2\right)+\left(\frac{12}{b+2}+b-2\right)+\left(\frac{12}{c+2}+c-2\right)\)
Mặt khác: \(\frac{12}{a+2}+a-2=\frac{12+a^2-4}{a+2}=\frac{a^2+8}{a+2}\)
Tương tự: \(\frac{12}{b+2}+b-2=\frac{b^2+8}{b+2}\); \(\frac{12}{c+2}+c-2=\frac{c^2+8}{c+2}\)
Từ đó suy ra \(a+b+c+6=\frac{a^2+8}{a+2}+\frac{b^2+8}{b+2}+\frac{c^2+8}{c+2}\)
\(\ge\frac{\left(\sqrt{a^2+8}+\sqrt{b^2+8}+\sqrt{c^2+8}\right)^2}{a+b+c+6}\)(Theo BĐT Bunyakovsky dạng phân thức)
\(\Rightarrow\left(a+b+c+6\right)^2\ge\left(\sqrt{a^2+8}+\sqrt{b^2+8}+\sqrt{c^2+8}\right)^2\)
hay \(\sqrt{a^2+8}+\sqrt{b^2+8}+\sqrt{c^2+8}\le a+b+c+6\)
Đẳng thức xảy ra khi a = b = c = 1
Ta có: \(a^2+ab+b^2\)
\(=\left(a+b\right)^2-ab\ge\left(a+b\right)^2-\frac{\left(a+b\right)^2}{4}=\frac{3\left(a+b\right)^2}{4}\)
\(\Rightarrow\sqrt{a^2+ab+b^2}\ge\sqrt{\frac{3\left(a+b\right)^2}{4}}=\frac{\sqrt{3}}{2}\left(a+b\right)\)
Tương tự, ta có: \(\sqrt{b^2+bc+c^2}\ge\frac{\sqrt{3}}{2}\left(b+c\right)\)
\(\sqrt{c^2+ca+a^2}\ge\frac{\sqrt{3}}{2}\left(c+a\right)\)
Do đó ta có: \(Q\ge\frac{\sqrt{3}}{2}\left(a+b+b+c+c+a\right)=\sqrt{3}\) ( Do a+b+c=1)
Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)
Để dễ nhìn, đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\)
\(VT=\frac{xy}{z^2+2xy}+\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}\)
\(2VT=\frac{2xy}{z^2+2xy}+\frac{2yz}{x^2+2yz}+\frac{2zx}{y^2+2xz}=1-\frac{z^2}{z^2+2xy}+1-\frac{x^2}{x^2+2yz}+1-\frac{y^2}{y^2+2xz}\)
\(2VT=3-\left(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\right)\)
\(2VT\le3-\frac{\left(x+y+z\right)^2}{x^2+2yz+y^2+2xz+z^2+2xy}=3-\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=2\)
\(\Rightarrow VT\le1\)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)
\(S=\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\\ =\sqrt{a^2+2ab+b^2-3ab}+\sqrt{b^2+2bc+c^2-3bc}+\sqrt{c^2+2ca+a^2-3ca}\\ =\sqrt{\left(a+b\right)^2-\dfrac{3}{4}\cdot4ab}+\sqrt{\left(b+c\right)^2-\dfrac{3}{4}\cdot4bc}+\sqrt{\left(c+a\right)^2-\dfrac{3}{4}\cdot4ca}\)
Áp dụng BDT : Cô-si:
\(\Rightarrow S=\sqrt{\left(a+b\right)^2-\dfrac{3}{4}\cdot4ab}+\sqrt{\left(b+c\right)^2-\dfrac{3}{4}\cdot4bc}+\sqrt{\left(c+a\right)^2-\dfrac{3}{4}\cdot4ca}\\ \ge\sqrt{\left(a+b\right)^2-\dfrac{3}{4}\cdot\left(a+b\right)^2}+\sqrt{\left(b+c\right)^2-\dfrac{3}{4}\left(b+c\right)^2}+\sqrt{\left(c+a\right)^2-\dfrac{3}{4}\left(c+a\right)^2}\\ =\sqrt{\dfrac{1}{4}\left(a+b\right)^2}+\sqrt{\dfrac{1}{4}\left(b+c\right)^2}+\sqrt{\dfrac{1}{4}\left(c+a\right)^2}\\ =\dfrac{1}{2}\left(a+b\right)+\dfrac{1}{2}\left(b+c\right)+\dfrac{1}{2}\left(c+a\right)\\ =\dfrac{1}{2}\left(a+b+b+c+c+a\right)\\ =a+b+c\\ =2019\)
Dấu "=" xảy ra khi:\(\left\{{}\begin{matrix}a=b=c\\a+b+c=2019\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=673\\b=673\\c=673\end{matrix}\right.\)
Vậy \(S_{Min}=2019\) khi \(a=b=c=673\)