1) Cho x > 1. Tìm GTNN của: \(A=\frac{1+x^4}{x\left(x-1\right)\left(x+1\right)}\)
2) Trong các cặp (x;y) thỏa mãn \(\frac{x^2-x+y^2-y}{x^2+y^2-1}\le0\). Tìm cặp có tổng x + 2y lớn nhất.
3) Cho x thỏa mãn \(x^2+\left(3-x\right)^2\ge5\). Tìm GTNN của \(A=x^4+\left(3-x\right)^4+6x^2\left(3-x\right)^2\)
4) Tìm GTNN của \(Q=\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)+\frac{1}{4}\left(x^{16}+y^{16}\right)-\left(1+x^2y^2\right)^2\)
5) Cho x, y > 1. Tìm GTNN của \(P=\frac{\left(x^3+y^3\right)-\left(x^2+y^2\right)}{\left(x-1\right)\left(y-1\right)}\)
6) Cho x, y, z > 0 thỏa mãn: \(xy^2z^2+x^2z+y=3z^2\). Tìm GTLN của \(P=\frac{z^4}{1+z^4\left(x^4+y^4\right)}\)
7) Cho a, b, c > 0. CMR:\(\frac{a^2}{b^2+c^2}+\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+b^2}\ge\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
8) Cho x>y>0. và \(x^5+y^5=x-y\). CMR: \(x^4+y^4<1\)
9) Cho \(1\le a,b,c\le2\). CMR: \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le10\)
10) Cho \(x,y,z\ge0\)CMR: \(\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}\le\sqrt[3]{\frac{x+y}{2}}+\sqrt[3]{\frac{y+z}{2}}+\sqrt[3]{\frac{z+x}{2}}\)
11) Cho \(x,y\ge0\)thỏa mãn \(x^2+y^2=1\)CMR: \(\frac{1}{\sqrt{2}}\le x^3+y^3\le1\)
12) Cho a,b,c > 0 và a + b + c = 12. CM: \(\sqrt{3a+2\sqrt{a}+1}+\sqrt{3b+2\sqrt{b}+1}+\sqrt{3c+2\sqrt{c}+1}\le3\sqrt{17}\)
13) Cho x,y,z < 0 thỏa mãn \(x+y+z\le\frac{3}{2}\). CMR: \(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\ge3\sqrt{17}\)
14) Cho a,b > 0. CMR: \(\left(\sqrt[6]{a}+\sqrt[6]{b}\right)\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\left(\sqrt{a}+\sqrt{b}\right)\le4\left(a+b\right)\)
15) Với a, b, c > 0. CMR: \(\frac{a^8+b^8+c^8}{a^3.b^3.c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
16) Cho x, y, z > 0 và \(x^3+y^3+z^3=1\)CMR: \(\frac{x^2}{\sqrt{1-x^2}}+\frac{y^2}{\sqrt{1-y^2}}+\frac{z^2}{\sqrt{1-z^2}}\ge2\)
a/ \(VT\ge\frac{\left(\sqrt{b}+\sqrt{c}\right)^2}{2\sqrt{a}}+\frac{\left(\sqrt{c}+\sqrt{a}\right)^2}{2\sqrt{b}}+\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\sqrt{c}}\)
\(VT\ge\frac{\left(\sqrt{b}+\sqrt{c}+\sqrt{c}+\sqrt{a}+\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}=2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
\(VT\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b}+\sqrt{c}\)
\(VT\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[3]{\sqrt{abc}}=\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
b/ \(VT=\sum\frac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}=\sum\frac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}\)
\(VT\le\sum\frac{x}{x+\sqrt{xz}+\sqrt{xy}}=\sum\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Bài 1 :
Áp dụng BĐT Cô - si cho 2 số không âm ta có :
\(VT=\Sigma_{cyc}\frac{b+c}{\sqrt{a}}\ge2\left(\Sigma_{cyc}\sqrt{\frac{bc}{a}}\right)\)
\(\Leftrightarrow\Sigma_{cyc}\frac{b+c}{\sqrt{a}}\ge\left(\sqrt{\frac{ca}{b}}+\sqrt{\frac{ab}{c}}\right)+\left(\sqrt{\frac{ab}{c}}+\sqrt{\frac{bc}{a}}\right)+\left(\sqrt{\frac{bc}{a}}+\sqrt{\frac{ca}{b}}\right)\)
\(\Leftrightarrow\Sigma_{cyc}\frac{b+c}{\sqrt{a}}\ge2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\ge\sqrt{a}+\sqrt{b}+\sqrt{c}\)
\(+3\sqrt[6]{abc}=\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)