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Từ giả thiết: \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7\Leftrightarrow\sqrt{c}=7-\sqrt{a}-\sqrt{b}\)
Xét hạng tử: \(\frac{1}{\sqrt{ab}+\sqrt{c}-6}=\frac{1}{\sqrt{ab}+7-\sqrt{a}-\sqrt{b}-6}=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}\)
Từ đó: \(N=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\frac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\frac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\sqrt{abc}-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-1}\)
\(=\frac{7-3}{3-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+7-1}=\frac{4}{9-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}\)
Mặt khác: \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2-\left(a+b+c\right)}{2}=13\)
Suy ra: \(N=\frac{4}{9-13}=-1\). Kết luận: N = -1.
Từ giả thiết: \sqrt{a}+\sqrt{b}+\sqrt{c}=7\Leftrightarrow\sqrt{c}=7-\sqrt{a}-\sqrt{b}a+b+c=7⇔c=7−a−b
Xét hạng tử: \frac{1}{\sqrt{ab}+\sqrt{c}-6}=\frac{1}{\sqrt{ab}+7-\sqrt{a}-\sqrt{b}-6}=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}ab+c−61=ab+7−a−b−61=(a−1)(b−1)1
Từ đó: N=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\frac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\frac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}N=(a−1)(b−1)1+(b−1)(c−1)1+(c−1)(a−1)1
=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\sqrt{abc}-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-1}=(a−1)(b−1)(c−1)a+b+c−3=abc−(ab+bc+ca)+(a+b+c)−1a+b+c−3
=\frac{7-3}{3-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+7-1}=\frac{4}{9-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}=3−(ab+bc+ca)+7−17−3=9−(ab+bc+ca)4
Mặt khác: \sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2-\left(a+b+c\right)}{2}=13ab+bc+ca=2(a+b+c)2−(a+b+c)=13
Suy ra: N=\frac{4}{9-13}=-1N=9−134=−1. Kết luận: N = -1.
Câu hỏi của hoàng thị huyền trang - Toán lớp 9 - Học toán với OnlineMath
Em tham khảo nhé!
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Ta có:\(H=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}+c}+\frac{\sqrt{b}-\sqrt{c}}{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}+a}+\frac{\sqrt{c}-\sqrt{a}}{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}+b}\)
\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{c}+\sqrt{a}\right)\left(\sqrt{c}+\sqrt{b}\right)}+\frac{\sqrt{b}-\sqrt{c}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}+\frac{\sqrt{c}-\sqrt{a}}{\left(\sqrt{b}+\sqrt{a}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(=\frac{a-b+b-c+c-a}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{c}+\sqrt{a}\right)}\)\(=0\)
Vậy \(H=0\)
Ta có:\(a^5+ab+b^2\ge3a^2b\)
Tương tự ta có:
\(VT\le\frac{1}{\sqrt{3ab\left(a+2c\right)}}+\frac{1}{\sqrt{3bc\left(b+2a\right)}}+\frac{1}{\sqrt{3ca\left(c+2b\right)}}\)
\(=\frac{1}{\sqrt{3}}\left(\sqrt{\frac{c}{c+2a}}+\sqrt{\frac{a}{b+2a}}+\sqrt{\frac{b}{2b+c}}\right)\)
Ta cũng có:\(a+2c=a+c+c\ge\frac{1}{3}\left(\sqrt{a}+2\sqrt{c}\right)^2\)
\(\Rightarrow VT\le\frac{\sqrt{c}}{\sqrt{a}+2\sqrt{c}}+\frac{\sqrt{a}}{\sqrt{b}+2\sqrt{a}}+\frac{\sqrt{b}}{\sqrt{c}+2\sqrt{b}}\)
Đặt \(x=\frac{\sqrt{a}}{\sqrt{c}};y=\frac{\sqrt{b}}{\sqrt{a}};z=\frac{\sqrt{c}}{\sqrt{b}};xyz=1\)
\(\Rightarrow VT\le\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\)
Giả sử \(xy\le1\) thì \(z\ge1\)
Ta có: \(\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=\frac{1}{2}\left(\frac{1}{\frac{x}{2}+1}+\frac{1}{\frac{y}{2}+1}\right)+\frac{1}{z+2}\)
\(\le\frac{1}{1\frac{\sqrt{xy}}{2}}+\frac{1}{z+2}\le1\)(Đpcm)
Dấu = khi \(a=b=c=1\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\) thì x, y, z > 0; x + y + z = 1. Quy về: \(\sqrt{\frac{1}{x}+\frac{1}{yz}}+\sqrt{\frac{1}{y}+\frac{1}{zx}}+\sqrt{\frac{1}{z}+\frac{1}{xy}}\ge\sqrt{\frac{1}{xyz}}+\sqrt{\frac{1}{x}}+\sqrt{\frac{1}{y}}+\sqrt{\frac{1}{z}}\)
\(\Leftrightarrow\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge1+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(\Leftrightarrow\frac{x}{\sqrt{x+yz}+\sqrt{yz}}+\frac{y}{\sqrt{y+zx}+\sqrt{zx}}+\frac{z}{\sqrt{z+xy}+\sqrt{xy}}\ge1\) (chuyển vế qua nhóm lại rồi liên hợp)
\(\Leftrightarrow\Sigma_{cyc}\frac{x}{\sqrt{x\left(x+y+z\right)+yz}+\sqrt{yz}}\ge1\Leftrightarrow\Sigma_{cyc}\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}+\sqrt{yz}}\ge1\)
BĐT này đúng! Thật vậy:
\(VT\ge\Sigma_{cyc}\frac{x}{\frac{\left(x+y\right)+\left(z+z\right)}{2}+\frac{\left(y+z\right)}{2}}=\Sigma_{cyc}\frac{x}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)
Ta có đpcm. Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\Leftrightarrow a=b=c=3\)