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\(\frac{1}{\sqrt{1+a^3}}=\frac{1}{\sqrt{\left(1+a\right)\left(a^2-a+1\right)}}\ge\frac{2}{a^2+2}\)
\(\Rightarrow VT\ge\frac{2}{a^2+2}+\frac{2}{b^2+2}+\frac{2}{c^2+2}\)
Đặt \(\left(a;b;c\right)=\left(2\sqrt{\frac{x}{y}};2\sqrt{\frac{y}{z}};2\sqrt{\frac{z}{x}}\right)\)
\(\Rightarrow VT\ge\frac{y}{2x+y}+\frac{z}{2y+z}+\frac{x}{2z+x}=\frac{y^2}{2xy+y^2}+\frac{z^2}{2yz+z^2}+\frac{x^2}{2zx+x^2}\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)
Dấu "=" xảy ra khi \(a=b=c=2\)
Lời giải:
Do $abc=1$ nên đặt:
\((\sqrt{a}, \sqrt{b}, \sqrt{c})=(\frac{x}{y}, \frac{y}{z}, \frac{z}{x})\) với $x,y,z>0$
Khi đó, bài toán trở thành: Cho $x,y,z>0$. CMR:
\(\frac{xz^2}{2z^2y+xy^2}+\frac{yx^2}{2x^2z+yz^2}+\frac{zy^2}{2y^2x+zx^2}\geq 1\)
Thật vậy, áp dụng BĐT Cauchy-Schwarz:
\(\frac{xz^2}{2z^2y+xy^2}+\frac{yx^2}{2x^2z+yz^2}+\frac{zy^2}{2y^2x+zx^2}=\frac{(xz)^2}{2xyz^2+(xy)^2}+\frac{(xy)^2}{2x^2yz+(yz)^2}+\frac{(yz)^2}{2xy^2z+(xz)^2}\)
\(\geq \frac{(xz+xy+yz)^2}{2xyz^2+(xy)^2+2x^2yz+(yz)^2+2xy^2z+(xz)^2}=\frac{(xy+yz+xz)^2}{(xy+yz+xz)^2}=1\)
Ta có đpcm.
Dấu "=" xảy ra khi $x=y=z$ hay $a=b=c=1$
Đặt \(\left(a;b;c\right)\rightarrow\left(\frac{2y'z'}{x'^2};\frac{2z'x'}{y'^2};\frac{2x'y'}{z'^2}\right)\) với x', y', z' > 0. Quy về chứng minh:
\(\Sigma_{cyc}\frac{x'^3}{\sqrt{x'^6+8y'^3z'^3}}\ge1\). Đặt \(\left(x'^3;y'^3;z'^3\right)=\left(x;y;z\right)\). Quy về:
\(\Sigma_{cyc}\frac{x}{\sqrt{x^2+8yz}}\ge1\). Đến đây em thấy khá quen thuộc, hình như là bài IMO nào đó, để tối lục lại.
3.
\(5a^2+2ab+2b^2=\left(a^2-2ab+b^2\right)+\left(4a^2+4ab+b^2\right)\)
\(=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)
Tương tự \(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c};\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\)
\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
\(\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)
\(=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{3}.\sqrt{3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}=\frac{\sqrt{3}}{3}\)
\(\Rightarrow MaxP=\frac{\sqrt{3}}{3}\Leftrightarrow a=b=c=\sqrt{3}\)
1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)
\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)
\(a^3+b^3+1=a^3+b^3+abc\ge ab\left(a+b+c\right)\)
=> \(\frac{\sqrt{1+a^3+b^3}}{ab}\ge\frac{\sqrt{ab\left(a+b+c\right)}}{ab}=\frac{\sqrt{a+b+c}}{\sqrt{ab}}\)
Tuong tu: \(\frac{\sqrt{1+b^3+c^3}}{bc}\ge\frac{\sqrt{a+b+c}}{\sqrt{bc}}\)
\(\sqrt{1+c^3+a^3}\ge\frac{\sqrt{a+b+c}}{\sqrt{ca}}\)
suy ra: \(\frac{\sqrt{1+a^3+b^3}}{ab}+\frac{\sqrt{1+b^3+c^3}}{bc}+\frac{\sqrt{1+c^3+a^3}}{ca}\ge\sqrt{a+b+c}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)\)
\(\ge\sqrt{3\sqrt[3]{abc}}.3\sqrt[3]{\frac{1}{\sqrt{ab}}.\frac{1}{\sqrt{bc}}.\frac{1}{\sqrt{ca}}}=3\sqrt{3}\) (dpcm)
ap dung bat dang thuc amgm
\(\sqrt{b^3+1}\) \(=\sqrt{\left(b+1\right)\left(b^2-b+1\right)}\le\frac{b+1+b^2-b+1}{2}\) \(=\frac{b^2+2}{2}\)
\(\Rightarrow\frac{a}{\sqrt{b^3+1}}\ge2.\frac{a}{b^2+2}\)
P=\(\frac{a}{\sqrt{b^3+1}}+\frac{b}{\sqrt{c^3+1}}+\frac{c}{\sqrt{a^3+1}}\ge2\left(\frac{a}{b^2+2}+\frac{b}{c^2+2}+\frac{c}{a^2+2}\right)\) \(\)
=\(2\left(\frac{a^2}{a\left(b^2+2\right)}+\frac{b^2}{b\left(c^2+2\right)}+\frac{c^2}{c\left(a^2+2\right)}\right)\)
tiep tuc ap dung bdt cauchy-swart dang phan thuc
\(\ge2\frac{\left(a+b+c\right)^2}{a\left(b^2+2\right)+b\left(c^2+2\right)+c\left(a^2+2\right)}\)=
\(\frac{1}{\sqrt{a^3+1}}=\frac{1}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}}\ge\frac{2}{a+1+a^2-a+1}=\frac{2}{a^2+2}\)
Thiết lập tương tự: \(\frac{1}{\sqrt{b^3+1}}\ge\frac{2}{b^2+2}\) ; \(\frac{1}{\sqrt{c^3+1}}\ge\frac{2}{c^2+2}\)
\(\Rightarrow VT\ge\frac{2}{a^2+2}+\frac{2}{b^2+2}+\frac{2}{c^2+2}=\frac{1}{\frac{a^2}{2}+1}+\frac{1}{\frac{b^2}{2}+1}+\frac{1}{\frac{c^2}{2}+1}\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xyz=\frac{1}{8}\)
\(\Rightarrow VT\ge\frac{x^2}{x^2+\frac{1}{2}}+\frac{y^2}{y^2+\frac{1}{2}}+\frac{z^2}{z^2+\frac{1}{2}}\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+\frac{3}{2}}\)
\(\Rightarrow VT\ge\frac{x^2+y^2+z^2+2\left(xy+yz+zx\right)}{x^2+y^2+z^2+\frac{3}{2}}\ge\frac{x^2+y^2+z^2+6.\sqrt[3]{\left(xyz\right)^2}}{x^2+y^2+z^2+\frac{3}{2}}=\frac{x^2+y^2+z^2+\frac{3}{2}}{x^2+y^2+z^2+\frac{3}{2}}=1\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\) hay \(a=b=c=2\)