1. Cho a,b,c > 0. Cmr: a) \(\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}+\frac{ab}{c^2+2ab}\le1\)
b) \(\frac{ab^2}{a^2+2b^2+c^2}+\frac{bc^2}{b^2+2c^2+a^2}+\frac{ca^2}{c^2+2a^2+b^2}\le\frac{a+b+c}{4}\)
2. Cho \(x,y,z>0;x+\frac{y}{3}+\frac{z}{5}\ge3;\frac{y}{3}+\frac{z}{5}\ge2;\frac{z}{5}\ge1.MaxP=x^2+y^2+z^2\)
3. Cho \(x>0;y\ge2;2x+y+xy\ge6.MinP=x^3+y^2\)
4. Cho \(0< \alpha< \beta< \gamma\). Giả sử x,y,z > 0 TM \(z\ge\gamma;\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}+\frac{xyz}{\alpha\beta\gamma}=4;\frac{y}{\beta}+\frac{z}{\gamma}+\frac{yz}{\beta\gamma}=3.MinP=x^3+y^3+z^3\)
Vì đã khuya nên não cũng không còn hoạt động tốt nữa, mình làm bài 1 thôi nhé.
Bài 1:
a)
\(2\text{VT}=\sum \frac{2bc}{a^2+2bc}=\sum (1-\frac{a^2}{a^2+2bc})=3-\sum \frac{a^2}{a^2+2bc}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\sum \frac{a^2}{a^2+2bc}\geq \frac{(a+b+c)^2}{a^2+2bc+b^2+2ac+c^2+2ab}=\frac{(a+b+c)^2}{(a+b+c)^2}=1\)
Do đó: \(2\text{VT}\leq 3-1\Rightarrow \text{VT}\leq 1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
b)
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\sum \frac{ab^2}{a^2+2b^2+c^2}=\sum \frac{ab^2}{\frac{a^2+b^2+c^2}{3}+\frac{a^2+b^2+c^2}{3}+\frac{a^2+b^2+c^2}{3}+b^2}\leq \sum \frac{1}{16}\left(\frac{9ab^2}{a^2+b^2+c^2}+\frac{ab^2}{b^2}\right)\)
\(=\frac{1}{16}.\frac{9(ab^2+bc^2+ca^2)}{a^2+b^2+c^2}+\frac{a+b+c}{16}(1)\)
Áp dụng BĐT AM-GM:
\(3(ab^2+bc^2+ca^2)\leq (a^2+b^2+c^2)(a+b+c)\)
\(\Rightarrow \frac{1}{16}.\frac{9(ab^2+bc^2+ca^2)}{a^2+b^2+c^2)}\leq \frac{3}{16}(a+b+c)(2)\)
Từ $(1);(2)\Rightarrow \text{VT}\leq \frac{a+b+c}{4}$ (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Bài 2/Áp dụng BĐT Bunyakovski:
\(\left(x^2+y^2+z^2\right)\left(1^2+3^2+5^2\right)\ge\left(x+3y+5z\right)^2\)
\(\Rightarrow P\ge\frac{\left(x+3y+5z\right)^2}{35}\) (*)
Ta có: \(x+3y+5z=x.1+\frac{y}{3}.9+\frac{z}{5}.25\)
\(=\frac{16z}{5}+8\left(\frac{y}{3}+\frac{z}{5}\right)+1\left(\frac{z}{5}+\frac{y}{3}+x\right)\)
\(\ge16+8.2+1.3=35\). Thay vào (*) là xong.
Đẳng thức xảy ra khi x = 1; y =3; z = 5
No choice teen, Akai Haruma, Arakawa Whiter, Phạm Lan Hương, soyeon_Tiểubàng giải, tth, Nguyễn Văn Đạt
giúp em với ạ! Cần gấp lắm! Thanks nhiều!
Ta có: \(b^2+c^2\ge2bc\)
\(\Rightarrow\frac{bc}{a^2+2bc}\le\frac{bc}{a^2+b^2+c^2}\)
Tương tự ta có:
\(\frac{ca}{b^2+2ca}\le\frac{ca}{a^2+b^2+c^2}\)
\(\frac{ab}{c^2+2ab}\le\frac{ab}{a^2+b^2+c^2}\)
\(\Rightarrow\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}+\frac{ab}{c^2+2ab}\le\frac{bc}{a^2+b^2+c^2}+\frac{ac}{a^2+b^2+c^2}+\frac{ab}{a^2+b^2+c^2}\)
\(\Rightarrow\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}+\frac{ab}{c^2+2ab}\le\frac{ab+bc+ca}{a^2+b^2+c^2}\)
\(\Rightarrow\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}+\frac{ab}{c^2+2ab}\le\frac{a^2+b^2+c^2}{a^2+b^2+c^2}\)
\(\Rightarrow\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}+\frac{ab}{c^2+2ab}\le1\left(đpcm\right)\)
2.
We have:
\(P=x^2+y^2+z^2=x\left(x+\frac{y}{3}+\frac{z}{5}\right)+\left(3y-x\right)\left(\frac{y}{3}+\frac{z}{5}\right)+\left(5z-3y\right)\frac{z}{5}\)\(\Rightarrow P\ge3x+2\left(3y-x\right)+\left(5z-3y\right)=x+3y+5z=\left(x+\frac{y}{3}+\frac{z}{5}\right)+8\left(\frac{y}{3}+\frac{z}{5}\right)+\frac{16z}{5}\)
\(\Rightarrow P\ge3+16+16=35\)
Dau '=' xay ra khi \(\left\{{}\begin{matrix}x=1\\y=3\\z=5\end{matrix}\right.\)
3) \(2P=\left(x^3+x^3+1\right)+2y^2-1\ge3x^2+2y^2-1\)
\(=\frac{3}{2}\left(x^2+1\right)+\frac{3}{2}\left(\frac{1}{4}y^2+1\right)+\frac{3}{2}\left(x^2+\frac{1}{4}y^2\right)+\frac{5}{4}y^2-4\)
\(\ge\frac{3}{2}\left(2x+y+xy\right)+\frac{5}{4}y^2-4\ge\frac{3}{2}.6+\frac{5}{4}.2^2-4=10\)
\(\Rightarrow\)\(P\ge5\)
"=" \(\Leftrightarrow\)\(x=1;y=2\)
