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\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2\)
\(\Leftrightarrow\frac{1}{1+a}=1-\frac{1}{1+b}+1-\frac{1}{1+c}\)
\(\Leftrightarrow\frac{1}{1+a}=\frac{b}{1+b}+\frac{c}{1+c}\ge2\sqrt{\frac{bc}{\left(1+b\right)\left(1+c\right)}}\left(\text{ta áp dụng BĐT cô-si}\right)\)
\(\frac{1}{1+b}\ge2\sqrt{\frac{ac}{\left(1+a\right)\left(1+c\right)}}\)
Tương tự, ta có:
\(\frac{1}{1+c}\ge2\sqrt{\frac{ac}{\left(1+a\right)\left(1+b\right)}}\)
Nhân theo vế. ta có: \(\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge8\frac{\sqrt{a^2b^2c^2}}{\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2}=\frac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)
\(\Leftrightarrow abc\le\frac{1}{8}\)
Dấu "=" xảy ra khi: \(Q=abc;MAX_Q=\frac{1}{8}\Leftrightarrow a=b=c=\frac{1}{2}\)
P/s: Ko chắc
Ta có: \(2a+b^2=2a\left(a+b+c\right)+b^2=b^2+2a^2+2ab+2ac\)
\(\ge4ab+2ac+a^2\)
\(\Rightarrow\frac{a}{2a+b^2}\le\frac{a}{4ab+2ac+a^2}=\frac{1}{4b+2c+a}\)
\(\le\frac{1}{49}.\frac{49}{4b+2c+a}=\frac{1}{49}.\frac{\left(4+2+1\right)^2}{4b+2c+a}\)
\(\le\frac{1}{49}\left(\frac{16}{4b}+\frac{4}{2c}+\frac{1}{a}\right)=\frac{1}{49}\left(\frac{4}{b}+\frac{2}{c}+\frac{1}{a}\right)\)
CMTT: \(\frac{b}{2b+c^2}\le\frac{1}{49}\left(\frac{4}{c}+\frac{2}{a}+\frac{1}{b}\right);\frac{c}{2c+a^2}\le\frac{1}{49}\left(\frac{4}{a}+\frac{2}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{a}{2a+b^2}+\frac{b}{2b+c^2}+\frac{c}{2c+a^2}\le\frac{1}{7}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)( đpcm )
MÌnh nghĩ đề phải là tìm GTLN chứ
Ta có: \(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}=2\)
\(\Rightarrow\frac{1}{a+b+1}=\frac{b+c}{b+c+1}+\frac{c+a}{c+a+1}\ge2\sqrt{\frac{\left(b+c\right)\left(c+a\right)}{\left(b+c+1\right)\left(c+a+1\right)}}\)
Tương tự: \(\frac{1}{b+c+1}\ge2\sqrt{\frac{\left(a+b\right)\left(c+a\right)}{\left(a+b+1\right)\left(c+a+1\right)}}\)
\(\frac{1}{c+a+1}\ge2\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{\left(a+b+1\right)\left(b+c+1\right)}}\)
Nhân lại ta có: \(\frac{1}{\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)}\ge\frac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\frac{1}{8}\)
Dấu = khi a=b=c=1/4
Mình có cách này,không chắc lắm:
\(VT=\frac{a}{a\left(a^2+bc+1\right)}+\frac{b}{b\left(b^2+ac+1\right)}+\frac{c}{c\left(c^2+ab+1\right)}\) (làm tắt,bạn tự hiểu nha)
\(=\frac{1}{a^2+bc+1}+\frac{1}{b^2+ac+1}+\frac{1}{c^2+ab+1}\)
\(\le\frac{1}{3}\left(\frac{1}{\sqrt[3]{a}}+\frac{1}{\sqrt[3]{b}}+\frac{1}{\sqrt[3]{c}}\right)\)
\(=\frac{1}{3}\left[\left(1+1+1\right)-\left(\frac{\sqrt[3]{a}-1}{\sqrt[3]{a}}+\frac{\sqrt[3]{b}-1}{\sqrt[3]{b}}+\frac{\sqrt[3]{c}-1}{\sqrt[3]{c}}\right)\right]\)
\(=1-\frac{1}{3}\left(\frac{\sqrt[3]{a}-1}{\sqrt[3]{a}}+\frac{\sqrt[3]{b}-1}{\sqrt[3]{b}}+\frac{\sqrt[3]{c}-1}{\sqrt[3]{c}}\right)\)
Áp dụng BĐT Cô si với biểu thức trong ngoặc:
\(=1-\frac{1}{3}\left(\frac{\sqrt[3]{a}-1}{\sqrt[3]{a}}+\frac{\sqrt[3]{b}-1}{\sqrt[3]{b}}+\frac{\sqrt[3]{c}-1}{\sqrt[3]{c}}\right)\)
\(\le1-\sqrt[3]{\left(\sqrt[3]{a}-1\right)\left(\sqrt[3]{b}-1\right)\left(\sqrt[3]{c-1}\right)}\le1^{\left(đpcm\right)}\)
Dấu "=" xảy ra khi a = b = c = 1
Ta c/m bđt sau:
\(a^3+1\ge a^2+a\)
\(\Leftrightarrow a^3+1-a^2-a\ge0\Leftrightarrow a\left(a^2-1\right)-\left(a^2-1\right)\ge0\Leftrightarrow\left(a-1\right)^2\left(a+1\right)\ge0\)
\(\Rightarrow\frac{a}{a^3+a+1}\le\frac{a}{a^2+2a}=\frac{1}{a+2}\)
\(\Rightarrow\frac{a}{a^3+a+1}+\frac{b}{b^3+b+1}+\frac{c}{c^3+c+1}\le\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\)
Đặt \((a,b,c)\rightarrow(\frac{x}{y},\frac{y}{z},\frac{z}{x})\)
\(\Rightarrow\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=\frac{y}{x+2y}+\frac{z}{y+2z}+\frac{x}{z+2x}=\frac{1}{2}\left(1-\frac{x}{x+2y}+1-\frac{y}{y+2z}+1-\frac{z}{z+2x}\right)=\frac{3}{2}-\frac{1}{2}\left(\frac{x^2}{x^2+2xy}+\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2xy}\right)\)\(\le\frac{3}{2}-\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}\right)=\frac{3}{2}-\frac{1}{2}.\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)
Dấu bằng xảy ra khi a=b=c=1
1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)
\(=ac+bc+c^2+ab\)
\(=a\left(b+c\right)+c\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\)
CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)
\(b+ca=\left(b+c\right)\left(a+b\right)\)
Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)
Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)
CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)
\(\Rightarrow P\le\frac{1}{2}.3\)
\(\Rightarrow P\le\frac{3}{2}\)
Dấu"="xảy ra \(\Leftrightarrow a=b=c\)
Vậy /...
\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)
\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)
Tương tự rồi cộng lại:
\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)
\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)
Dấu "=" xảy ra tại \(a=b=c=1\)
Sửa đề: \(\frac{a}{b}+\frac{a}{c}+\frac{c}{b}+\frac{c}{a}+\frac{b}{c}+\frac{b}{a}\ge\sqrt{2}\left(\Sigma\sqrt{\frac{1-a}{a}}\right)\)
or \(\Sigma\frac{b+c}{a}\ge\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}\)
Theo AM-GM:\(\frac{b+c}{a}\ge2\sqrt{\frac{2\left(b+c\right)}{a}}-2\)
Tương tự và cộng lại: \(VT\ge2\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}-6\)
Mà: \(\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}\ge3\sqrt[6]{\frac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}}\ge6\)
Từ đó: \(VT\ge2\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}-\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}=VP\)
Done!
Gợi ý này: Đặt \(a=x^3,b=y^3,c=z^3\) rồi áp dụng bất đẳng thức này \(x^3+y^3\ge xy\left(x+y\right)\) rồi biến đổi 1 chút nx là ra