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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 \(a=b=c=\frac{1}{3}\)
a/ \(\frac{b}{b}.\sqrt{\frac{a^2+b^2}{2}}+\frac{c}{c}.\sqrt{\frac{b^2+c^2}{2}}+\frac{a}{a}.\sqrt{\frac{c^2+a^2}{2}}\)
\(\le\frac{1}{b}.\left(\frac{3b^2+a^2}{4}\right)+\frac{1}{c}.\left(\frac{3c^2+b^2}{4}\right)+\frac{1}{a}.\left(\frac{3a^2+c^2}{4}\right)\)
\(=\frac{1}{4}.\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+\frac{3}{4}.\left(a+b+c\right)\)
Ta cần chứng minh
\(\frac{1}{4}.\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+\frac{3}{4}.\left(a+b+c\right)\le\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
\(\Leftrightarrow\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\ge\left(a+b+c\right)\)
Mà: \(\Leftrightarrow\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\ge\frac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)
Vậy có ĐPCM.
Câu b làm y chang.
Bạn tham khảo:
Câu hỏi của Phạm Vũ Trí Dũng - Toán lớp 8 | Học trực tuyến
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
\(S=\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\\ =\sqrt{a^2+2ab+b^2-3ab}+\sqrt{b^2+2bc+c^2-3bc}+\sqrt{c^2+2ca+a^2-3ca}\\ =\sqrt{\left(a+b\right)^2-\dfrac{3}{4}\cdot4ab}+\sqrt{\left(b+c\right)^2-\dfrac{3}{4}\cdot4bc}+\sqrt{\left(c+a\right)^2-\dfrac{3}{4}\cdot4ca}\)
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 \(a=b=c=673\)