Bài 1: Cho các số thực dương a,b,c thỏa mãn các điều kiện \(\left(a+c\right)\left(b+c\right)=4c^2\). Tìm giá trị lớn nhất, giá trị nhỏ nhất của biểu thức
\(P=\frac{a}{b+3c}+\frac{b}{a+3c}+\frac{ab}{bc+ca}\)
Bài 2: Cho x,y,z thỏa mãn x+y+z=0 và \(x^2+y^2+z^2=1\). Tìm GTLN của biểu thức \(P=x^5+y^5+z^5\)
Bài 3: Cho a,b,c dương thỏa mãn \(a+b+c=1.\)Tìm Min
\(P=2020\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+\frac{1}{3\left(a^2+b^2+c^2\right)}\)
Bài 4: Cho a,b,c là các số thực không âm thỏa mãn điều kiện a+b+c=3. Tìm GTLN của biểu thức \(P=a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1}\)
1.
\(\left(a+c\right)\left(b+c\right)=4c^2\Leftrightarrow\left(\frac{a}{c}+1\right)\left(\frac{b}{c}+1\right)=4\)
Đặt \(\left(\frac{a}{c};\frac{b}{c}\right)=\left(x;y\right)\Rightarrow\left(x+1\right)\left(y+1\right)=4\)
\(\Leftrightarrow xy+x+y=3\Rightarrow\left\{{}\begin{matrix}xy=3-\left(x+y\right)\\4\le\frac{1}{4}\left(x+y+2\right)^2\Rightarrow x+y\ge2\end{matrix}\right.\)
Ta có:
\(P=\frac{\frac{a}{c}}{\frac{b}{c}+3}+\frac{\frac{b}{c}}{\frac{a}{c}+3}+\frac{1}{\frac{c}{a}+\frac{c}{b}}=\frac{x}{y+3}+\frac{y}{x+3}+\frac{1}{\frac{1}{x}+\frac{1}{y}}=\frac{x}{y+3}+\frac{y}{x+3}+\frac{xy}{x+y}\)
\(P=\frac{x^2+y^2+3\left(x+y\right)}{\left(x+1\right)\left(y+1\right)+2\left(x+y\right)+8}+\frac{xy}{x+y}=\frac{\left(x+y\right)^2-2xy+3\left(x+y\right)}{2\left(x+y\right)+12}+\frac{xy}{x+y}\)
Đặt \(x+y=t\Rightarrow2\le t< 3\Rightarrow P=\frac{t^2-2\left(3-t\right)+3t}{2t+12}+\frac{3-t}{t}=\frac{t^2+5t-6}{2\left(t+6\right)}+\frac{3}{t}-1\)
\(P=\frac{\left(t-1\right)\left(t+6\right)}{2\left(t+6\right)}+\frac{3}{t}-1=\frac{t}{2}+\frac{3}{t}-\frac{1}{2}\)
\(P\ge2\sqrt{\frac{3t}{2t}}-\frac{1}{2}=\sqrt{6}-\frac{1}{2}\)
\(P_{min}=\sqrt{6}-\frac{1}{2}\) khi \(t=\sqrt{6}\)
\(P=\frac{t}{2}+\frac{3}{t}-\frac{1}{2}=\frac{t^2-5t+6}{2t}+2=\frac{\left(t-2\right)\left(t-3\right)}{2t}+2\le2\)
\(P_{max}=2\) khi \(t=2\)
2.
\(xy+yz+zx=\frac{\left(x+y+z\right)^2-\left(x^2+y^2+z^2\right)}{2}=-\frac{1}{2}\)
\(\Rightarrow yz=-\frac{1}{2}-x\left(y+z\right)=-\frac{1}{2}-x\left(-x\right)=x^2-\frac{1}{2}\)
Ta có:
\(x+y=-z\Leftrightarrow\left(x+y\right)^5=-z^5\)
\(\Leftrightarrow x^5+y^5+z^5=-5x^4y-10x^3y^2-10x^2y^3-5xy^4\)
\(\Leftrightarrow x^5+y^5+z^5=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(\Leftrightarrow P=-5xy\left[\left(x+y\right)^3-xy\left(x+y\right)\right]\)
\(\Leftrightarrow P=-5xy\left[-z^3+xyz\right]=5xyz\left(z^2-xy\right)\)
\(\Leftrightarrow P=\frac{5}{2}xyz\left(z^2+\left(x+y\right)^2-2xy\right)=\frac{5}{2}xyz\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow P=\frac{5}{2}xyz=\frac{5}{2}x\left(x^2-\frac{1}{2}\right)\)
\(\Rightarrow P^2=\frac{25}{4}x^2\left(\frac{1}{2}-x^2\right)^2=\frac{25}{8}.2x^2\left(\frac{1}{2}-x^2\right)\left(\frac{1}{2}-x^2\right)\)
\(\Rightarrow P^2\le\frac{25}{8}\left(\frac{2x^2+\frac{1}{2}-x^2+\frac{1}{2}-x^2}{3}\right)^3=\frac{25}{216}\)
\(\Rightarrow P\le\frac{5\sqrt{6}}{36}\)
\(P_{max}=\frac{5\sqrt{6}}{36}\) khi \(x=-\frac{1}{\sqrt{6}}\)
3.
Xét \(Q=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
\(Q^2=\frac{a^4}{b^2}+\frac{2a^2b}{c}+c^2+\frac{b^4}{c^2}+\frac{2b^2c}{a}+a^2+\frac{c^4}{a^2}+\frac{2c^2a}{b}+b^2-\left(a^2+b^2+c^2\right)\)
\(\Rightarrow Q^2\ge4\sqrt[4]{\frac{a^4.a^2b.a^2b.c^2}{b^2c^2}}+4\sqrt[4]{\frac{b^4.b^2c.c^2c.a^2}{c^2a^2}}+4\sqrt[4]{\frac{c^4.c^2a.c^2a.b^2}{a^2b^2}}-\left(a^2+b^2+c^2\right)\)
\(\Rightarrow Q^2\ge3\left(a^2+b^2+c^2\right)\Rightarrow Q\ge\sqrt{3\left(a^2+b^2+c^2\right)}\)
Đặt \(x=a^2+b^2+c^2\ge\frac{1}{3}\)
\(\Rightarrow P\ge2020\sqrt{3x}+\frac{1}{3x}=\sqrt{3x}+\sqrt{3x}+\frac{1}{3x}+2018\sqrt{3x}\)
\(\Rightarrow P\ge3\sqrt[3]{\frac{3x}{3x}}+2018.\sqrt{3.\frac{1}{3}}=2021\)
\(P_{min}=2021\) khi \(a=b=c=\frac{1}{3}\)
4.
\(a\sqrt{b^3+1}=a\sqrt{\left(b+1\right)\left(b^2-b+1\right)}\le\frac{1}{2}a\left(b+1+b^2-b+1\right)=a+\frac{ab^2}{2}\)
Tương tự: \(b\sqrt{c^3+1}\le b+\frac{bc^2}{2}\) ; \(c\sqrt{a^3+1}\le c+\frac{ca^2}{2}\)
\(\Rightarrow P\le a+b+c+\frac{1}{2}\left(a^2c+b^2a+c^2b\right)=3+\frac{1}{2}\left(a^2c+b^2a+c^2b\right)\)
Xét \(Q=a^2c+b^2a+c^2b\le a^2c+b^2a+c^2b+abc\)
Không mất tính tổng quát, giả sử \(a=mid\left\{a;b;c\right\}\)
\(\Rightarrow\left(a-b\right)\left(a-c\right)\le0\)
\(\Leftrightarrow a^2+bc\le ab+ac\)
\(\Leftrightarrow a^2c+c^2b\le abc+ac^2\)
\(\Rightarrow Q\le b^2a+abc+abc+ac^2=a\left(b+c\right)^2\)
\(\Rightarrow Q\le\frac{1}{2}.2a\left(b+c\right)\left(b+c\right)\le\frac{1}{2}\left(\frac{2a+b+c+b+c}{3}\right)^3=4\)
\(Q_{max}=4\) khi \(\left(a;b;c\right)=\left(1;2;0\right)\) và 1 số hoán vị
Nguyễn Việt Lâm cảm ơn anh nha
Anh có thể giúp em mấy bài nữa ko anh vào câu hỏi của em í ạ
