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Bài này cũng dễ mà:
Áp dụng BĐT Cô-si, ta có:
\(y+z+1\ge3\sqrt[3]{yz}\)
\(\Rightarrow\)\(\dfrac{y+z+1}{3}\ge\sqrt[3]{yz}\)
\(\Rightarrow\)\(\dfrac{x}{\sqrt[3]{yz}}\ge\dfrac{3x}{y+z+1}\)
\(\Rightarrow\)\(\sum\dfrac{x}{\sqrt[3]{yz}}\ge\sum\dfrac{3x}{y+z+1}\)
Mà \(\sum\dfrac{3x}{y+z+1}=\sum\dfrac{3x^2}{xy+xz+x}\)
Áp dụng BĐT Cauchy -Schwaz:
\(\sum\dfrac{3x^2}{xy+xz+x}\ge\dfrac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\)
Mà:
\(xy+yz+xz\le x^2+y^2+z^2\)(BĐT phụ)
\(\Rightarrow\)\(2\left(xy+yz+xz\right)\le2\left(x^2+y^2+z^2\right)=6\)
Áp dụng BĐT Bunhicopski:
\(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\)
\(\Rightarrow x+y+z\le3\)
\(\Rightarrow2\left(xy+yz+xz\right)+x+y+z\le6+3=9\)
\(\Rightarrow\)\(\dfrac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{3\left(x+y+z\right)^2}{9}\ge\dfrac{\left(x+y+z\right)^2}{3}\ge xy+yz+xz\left(ĐPCM\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\)x=y=z=1
Ta có \(x^3+y^3\ge xy\left(x+y\right)\)
\(\Rightarrow1+x^3+y^3\ge xyz+xy\left(x+y+z\right)=xy\left(x+y+z\right)\)
Tương tự ta có
\(VT\ge\dfrac{\sqrt{xy\left(x+y+z\right)}}{xy}+\dfrac{\sqrt{yz\left(x+y+z\right)}}{yz}+\dfrac{\sqrt{zx\left(x+y+z\right)}}{zx}\)
\(=\sqrt{x+y+z}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\)
\(=\sqrt{x+y+z}.\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{xyz}}\)
\(\ge\sqrt{3\sqrt[3]{xyz}}.\dfrac{3\sqrt[6]{xyz}}{1}=3\sqrt{3}\)
\("="\Leftrightarrow x=y=z=1\)
Ta xét BĐT phụ: \(1+x^3+y^3\ge xy\left(x+y+z\right)\)
\(x^3+y^3\ge xy\left(x+y\right)+xyz-1\)
\(x^3+y^3-xy\left(x+y\right)\ge0\)
\(\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\ge0\)
\(\left(x+y\right)\left(x-y\right)^2\ge0\)( Luôn đúng, vậy BĐT phụ đúng)
\(\sum\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\sum\dfrac{\sqrt{xy\left(x+y+z\right)}}{xy}=\sqrt{x+y+z}.\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\ge\sqrt{3\sqrt[3]{xyz}}.\left(3\sqrt[3]{\dfrac{1}{\sqrt{x^2y^2z^2}}}\right)=3\sqrt{3}\)
GTNN của P là \(3\sqrt{3}\Leftrightarrow x=y=z=1\)
Gọi \(A=\sum\dfrac{x^3}{\sqrt{y^2+3}}\)
Theo Holder: \(A.A.\left(\left(y^2+3\right)+\left(z^2+3\right)+\left(x^2+3\right)\right)\ge\left(x^3+y^3+z^3\right)^3\)
\(\Rightarrow A^2\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{x^2+y^2+z^2+9}\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}=\dfrac{\left(x^3+y^3+z^3\right)^3}{\left(x+y+z\right)^2+xy+yz+zx}\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{\left(x+y+z\right)^2+\dfrac{\left(x+y+z\right)^2}{3}}\)
Ta có đánh giá sau: \(x^3+y^3+z^3\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x+y+z}\ge\dfrac{\left(x+y+z\right)^3}{9}\)
\(\Rightarrow A^2\ge\dfrac{\dfrac{\left(x+y+z\right)^3}{9}}{\left(x+y+z\right)^2+\dfrac{\left(x+y+z\right)^2}{3}}=\dfrac{x+y+z}{12}\ge\dfrac{\sqrt{3\left(xy+yz+zx\right)}}{12}\ge\dfrac{1}{4}\)
\(\Rightarrow A\ge\dfrac{1}{2}\)
Bài 1:
Ta có: \(\dfrac{2a}{\sqrt{1+a^2}}=\dfrac{2a}{\sqrt{ab+bc+ca+a^2}}=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\dfrac{b}{\sqrt{1+b^2}}=\dfrac{b}{\sqrt{ab+bc+ca+b^2}}=\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}\)
\(\dfrac{c}{\sqrt{1+c^2}}=\dfrac{c}{\sqrt{ab+bc+ca+c^2}}=\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
Vậy \(P=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
Áp dụng BĐT AM-GM ta có:
\(P\le a\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+b\left(\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{a+c}\right)+c\left(\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{a+c}\right)=\dfrac{9}{4}\)
Bài 2:
Ta có:
\(\dfrac{1+\sqrt{1+x^2}}{x}=\dfrac{2+\sqrt{4\left(1+x^2\right)}}{2x}\le\dfrac{2+\dfrac{4+\left(1+x^2\right)}{2}}{2x}=\dfrac{9+x^2}{4x}\)
Tương tự ta cũng có:
\(\dfrac{1+\sqrt{1+y^2}}{y}\le\dfrac{9+y^2}{4y};\dfrac{1+\sqrt{1+z^2}}{z}\le\dfrac{9+z^2}{4z}\)
Cộng theo vế 3 BĐT trên ta có:
\(\dfrac{1+\sqrt{1+x^2}}{x}+\dfrac{1+\sqrt{1+y^2}}{y}+\dfrac{1+\sqrt{1+z^2}}{z}\le\dfrac{9+x^2}{4x}+\dfrac{9+y^2}{4y}+\dfrac{9+z^2}{4z}\)
\(=\dfrac{9\left(xy+yz+xz\right)+xyz\left(x+y+z\right)}{4xyz}\le\dfrac{9\cdot\dfrac{\left(x+y+z\right)^2}{3}+\left(xyz\right)^2}{4xyz}=xyz\)
Đẳng thức xảy ra khi \(x=y=z=\sqrt{3}\)
Bài 1:
\(\dfrac{2a}{\sqrt{1+a^2}}=\dfrac{2a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
Sau đó côsi
Tự làm nốt nhé, ra 3/2 đấy. Em học lớp 8 nên cách giải chỉ thế thôi. Câu 2 em chưa làm được