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Câu hỏi của Lê Đình Quân - Toán lớp 9 | Học trực tuyến
Đặt vế trái là P và \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=4\)
Ta cần chứng minh: \(P=\frac{1}{xy+2yz+zx}+\frac{1}{xy+yz+2zx}+\frac{1}{2xy+yz+zx}\le\frac{1}{xyz}\)
\(P=\frac{1}{xy+yz+yz+zx}+\frac{1}{xy+yz+zx+zx}+\frac{1}{xy+xy+yz+zx}\)
\(P\le\frac{1}{16}\left(\frac{1}{xy}+\frac{2}{yz}+\frac{1}{zx}+\frac{1}{xy}+\frac{1}{yz}+\frac{2}{zx}+\frac{2}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)
\(P\le\frac{1}{4}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\frac{1}{4}\left(\frac{x+y+z}{xyz}\right)=\frac{1}{4}.\frac{4}{xyz}=\frac{1}{xyz}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=\frac{4}{3}\) hay \(a=b=c=\frac{16}{9}\)
\(VT=\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{1}{\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{2}{\sqrt{c}}}\right)\le\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{\sqrt{a}+\sqrt{b}+2\sqrt{c}}{16}\right)=\frac{1}{\sqrt{abc}}\)
Dấu "=" xay ra khi \(a=b=c=\frac{16}{9}\)
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Sửa đề:\(\sqrt{\frac{ab}{a+b+1}}+\sqrt{\frac{b}{bc+c+1}}+\sqrt{\frac{a}{ca+c+1}}\ge\sqrt{3}\)Giả thiết \(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+c=3\) (chia hai vế của giả thiết ab > 0)
Hay \(x+y+z=3\left(\text{với }x=\frac{1}{a};y=\frac{1}{b};z=c\right)\)
Khi đó BĐT quy về: \(\sqrt{\frac{1}{x+y+xy}}+\sqrt{\frac{1}{y+z+yz}}+\sqrt{\frac{1}{z+x+zx}}\ge\sqrt{3}\)
Áp dụng trực tiếp BĐT AM-GM cho 3 số:
\(VT\ge3\sqrt[6]{\frac{1}{\left(x+y+xy\right)\left(y+z+yz\right)\left(z+x+zx\right)}}\)
\(=\frac{3\sqrt[6]{3^3}}{\sqrt[6]{\left(x+y+xy\right)\left(y+z+yz\right)\left(z+x+zx\right).3.3.3}}\)
\(=\frac{3\sqrt{3}}{\sqrt[6]{\left(x+y+xy\right)\left(y+z+yz\right)\left(z+x+zx\right).3.3.3}}\)
\(\ge\frac{18\sqrt{3}}{2\left(x+y+z\right)+xy+yz+zx+9}\)
\(\ge\frac{18}{2\left(x+y+z\right)+\frac{\left(x+y+z\right)^2}{3}+9}=\sqrt{3}\)
Check hộ em thử xem sửa đề có đúng không:D Thấy đề nó sai sai nên em sửa thôi:) Với lại đang buồn ngủ nên em chả biết có ngược dấu chỗ nào chăng@@
Từ giả thiết ta suy ra
\(\dfrac{1}{a}+\dfrac{1}{b}+c=3\)
Đặt \(\left(x;y;z\right)=\left(\dfrac{1}{a};\dfrac{1}{b};c\right)\Rightarrow x+y+z=3\)
\(VT=\dfrac{1}{\sqrt{xy+x+y}}+\dfrac{1}{\sqrt{yz+y+z}}+\dfrac{1}{\sqrt{xz+x+z}}\)
Ta chứng minh: \(\left(x+1+y\right)^2\ge3\left(xy+x+y\right)\)(Luôn đúng)
\(\Rightarrow VT\ge\dfrac{\sqrt{3}}{x+y+1}+\dfrac{\sqrt{3}}{y+z+1}+\dfrac{\sqrt{3}}{z+x+1}\ge\dfrac{9\sqrt{3}}{2\left(x+y+z\right)+3}=\sqrt{3}\)
a) Ta có BĐT:
\(a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\ge\left(a+b\right)ab\)
\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b+c\right)\)
\(\Rightarrow\frac{1}{a^3+b^3+abc}\le\frac{1}{ab\left(a+b+c\right)}\)
Tương tự cho 2 bất đẳng thức còn lại rồi cộng theo vế:
\(VT\le\frac{1}{ab\left(a+b+c\right)}+\frac{1}{bc\left(a+b+c\right)}+\frac{1}{ca\left(a+b+c\right)}\)
\(=\frac{a+b+c}{abc\left(a+b+c\right)}=\frac{1}{abc}=VP\)
Khi \(a=b=c\)