Bài 1: Cho a,b,c dương
a) Tìm Max \(P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\)
b) Tìm Max \(Q=\frac{a^2}{5a^2+\left(b+c\right)^2}+\frac{b^2}{5b^2+\left(c+a\right)^2}+\frac{c^2}{5c^2+\left(a+b\right)^2}\)
Bài 2: Cho x,y,z là các số thực không âm thỏa mãn \(x+y+z=\frac{3}{2}\).Chứng minh rằng \(x+2xy+4xyz\le2\)
Bài 3: Cho a,b thỏa mãn \(\left(x+y\right)^3+4xy\ge2\). Tìm Min \(P=3\left(x^4+y^4+x^2y^2\right)-2\left(x^2+y^2\right)+1\)
Bài 4: Cho x,y,z >0: \(x\left(x+y+z\right)=3yz\). Chứng minh: \(\left(x+y\right)^3+\left(x+z\right)^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\le5\left(y+z\right)^3\)
Bài 5:Cho a,b,c không âm thỏa mãn \(a^2+b^2+c^2+abc=4\). CMR: \(a+b+c\le3\)
1a.
\(2P=1-\frac{bc}{2a^2+bc}+1-\frac{ca}{2b^2+ca}+1-\frac{ab}{2c^2+ab}\)
\(\Rightarrow2P=3-\left(\frac{bc}{2a^2+bc}+\frac{ca}{2b^2+ca}+\frac{ab}{2c^2+ab}\right)\)
\(\Rightarrow2P=3-\left(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{c^2a^2}{2b^2ca+c^2a^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\right)\)
\(\Rightarrow2P\le3-\frac{\left(ab+bc+ca\right)^2}{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)}=3-1=2\)
\(\Rightarrow P\le1\)
\(P_{max}=1\) khi \(a=b=c\)
1b.
\(Q=\frac{a^2}{5a^2+b^2+c^2+2bc}+\frac{b^2}{5b^2+a^2+c^2+2ca}+\frac{c^2}{5c^2+a^2+b^2+2ab}\)
\(Q=\frac{a^2}{a^2+b^2+c^2+\left(2a^2+bc\right)+\left(2a^2+bc\right)}+\frac{b^2}{a^2+b^2+c^2+\left(2b^2+ca\right)+\left(2b^2+ca\right)}+\frac{c^2}{a^2+b^2+c^2+\left(2c^2+ab\right)+\left(2c^2+ab\right)}\)
\(\Rightarrow Q\le\frac{1}{9}\left(\frac{a^2}{a^2+b^2+c^2}+\frac{b^2}{a^2+b^2+c^2}+\frac{c^2}{a^2+b^2+c^2}+2\left(\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\right)\right)\)
\(\Rightarrow Q\le\frac{1}{9}\left(1+2\left(\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\right)\right)\)
Theo kết quả câu a ta có:
\(\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\le1\)
\(\Rightarrow Q\le\frac{1}{9}\left(1+2\right)=\frac{1}{3}\)
\(Q_{max}=\frac{1}{3}\) khi \(a=b=c\)
2.
Do \(x+y+z=\frac{3}{2}\Rightarrow x< \frac{3}{2}< 2\)
Ta có:
\(VT=x+x.4y\left(\frac{1}{2}+z\right)\le x+x\left(\frac{1}{2}+z+y\right)^2=x+x\left(\frac{1}{2}+\frac{3}{2}-x\right)^2\)
\(\Leftrightarrow VT\le x+x\left(2-x\right)^2=x^3-4x^2+5x\)
\(\Leftrightarrow VT\le x^3-4x^2+5x-2+2=\left(x-1\right)^2\left(x-2\right)+2\le2\) (đpcm)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;\frac{1}{2};0\right)\)
3.
Ta có:
\(\left(x+y\right)^3+4xy\ge2\Leftrightarrow\left(x+y\right)^3+\left(x+y\right)^2-\left(x-y\right)^2\ge2\)
\(\Leftrightarrow\left(x+y\right)^3+\left(x+y\right)^2\ge2+\left(x-y\right)^2\ge2\)
\(\Leftrightarrow\left(x+y\right)^3+\left(x+y\right)^2-2\ge0\)
\(\Leftrightarrow\left(x+y-1\right)\left[\left(x+y+1\right)^2+1\right]\ge0\)
\(\Leftrightarrow x+y\ge1\Rightarrow x^2+y^2\ge\frac{1}{2}\)
Do đó:
\(P=\frac{3}{2}\left(x^2+y^2\right)^2+\frac{3}{2}\left(x^4+y^4\right)-2\left(x^2+y^2\right)+1\)
\(P\ge\frac{3}{2}\left(x^2+y^2\right)+\frac{3}{4}\left(x^2+y^2\right)^2-2\left(x^2+y^2\right)+1\)
\(P\ge\frac{9}{4}\left(x^2+y^2\right)^2-2\left(x^2+y^2\right)+1\)
Đặt \(x^2+y^2=a\ge\frac{1}{2}\)
\(P\ge\frac{9}{4}a^2-2a+1=\frac{9}{4}a^2-2a+\frac{7}{16}+\frac{9}{16}\)
\(P\ge\frac{9}{4}\left(a-\frac{1}{2}\right)\left(a-\frac{7}{18}\right)+\frac{9}{16}\ge\frac{9}{16}\)
\(P_{min}=\frac{9}{16}\) khi \(a=\frac{1}{2}\) hay \(x=y=\frac{1}{2}\)
4.
\(x^2+xy+yz+zx=4yz\Leftrightarrow\left(x+y\right)\left(x+z\right)=4yz\)
Ta có:
\(\left(x+y\right)^3+\left(x+z\right)^3=\left(x+y+x+z\right)\left[\left(x+y-\left(x+z\right)\right)^2+\left(x+y\right)\left(x+z\right)\right]\)
\(=\left(x+y+x+z\right)\left[\left(y-z\right)^2+4yz\right]=\left(x+y+x+z\right)\left(y+z\right)^2\)
Lại có:
\(x+y+x+z\le\sqrt{2\left[\left(x+y\right)^2+\left(x+z\right)^2\right]}=\sqrt{2\left[\left(y-z\right)^2+2\left(x+y\right)\left(x+z\right)\right]}\)
\(\Rightarrow x+y+x+z\le\sqrt{2\left[\left(y-z\right)^2+8yz\right]}\le\sqrt{2\left[2\left(y-z\right)^2+8yz\right]}\)
\(\Rightarrow x+y+x+z\le\sqrt{4\left(y+z\right)^2}=2\left(y+z\right)\)
\(\Rightarrow\left(x+y\right)^3+\left(x+z\right)^3\le2\left(y+z\right)^3\)
Do đó ta chỉ cần chứng minh:
\(3\left(x+y\right)\left(y+z\right)\left(x+z\right)\le3\left(y+z\right)^3\)
\(\Leftrightarrow\left(x+y\right)\left(x+z\right)\le\left(y+z\right)^2\)
Điều này hiển nhiên đúng do \(\left(x+y\right)\left(x+z\right)=4yz\le\left(y+z\right)^2\)
Dấu "=" xảy ra khi \(x=y=z\)
5.
Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có 2 số cùng phía so với 1
Không mất tính tổng quát, giả sử đó là a và b
\(\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\)
\(\Leftrightarrow ab+1\ge a+b\)
\(\Leftrightarrow2abc+2c\ge2ac+2bc\)
\(\Leftrightarrow a^2+b^2+c^2+2abc+1\ge a^2+b^2+c^2+2ac+2bc-2c+1\)
Mà \(a^2+b^2+c^2+1-2c+2ac+2bc\ge2ab+2c-2c+2ac+2bc\)
\(\Rightarrow a^2+b^2+c^2+2abc+1\ge2\left(ab+bc+ca\right)\)
Từ giả thiết ta có:
\(2a^2+2b^2+2c^2+2abc=8\)
\(\Leftrightarrow9=\left(a^2+b^2+c^2+2abc+1\right)+a^2+b^2+c^2\ge2\left(ab+bc+ca\right)+a^2+b^2+c^2\)
\(\Rightarrow9\ge\left(a+b+c\right)^2\)
\(\Rightarrow a+b+c\le3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Nguyễn Việt Lâm Cảm ơn anh nha
