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Đặt VT là K.
Ta có: \(6a^2+8ab+11b^2=\left(2a+3b\right)^2+2\left(a-b\right)^2\ge\left(2a+3b\right)^2\)
\(\Rightarrow\frac{a^2+3ab+b^2}{\sqrt{6a^2+8ab+11b^2}}\le\frac{a^2+3ab+b^2}{2a+3b}\)
Tiếp tục ta chứng minh: \(\frac{a^2+3ab+b^2}{2a+3b}\le\frac{3a+2b}{5}\Leftrightarrow\left(a-b\right)^2\ge0\)(đúng)
Tương tự ta có: \(\frac{b^2+3bc+c^2}{\sqrt{6b^2+8bc+11c^2}}\le\frac{3b+2c}{5}\);\(\frac{c^2+3ca+a^2}{\sqrt{6c^2+8ca+11a^2}}\le\frac{3c+2a}{5}\)
Cộng từng vế của các bđt trên, ta được:
\(M\le\frac{3b+2c}{5}+\frac{3a+3b}{5}+\frac{3c+2a}{5}=a+b+c\)
Lại có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\le a^2+b^2+c^2+\left(a^2+b^2\right)+\left(b^2+c^2\right)+\left(c^2+a^2\right)\)
hay \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)=9\Rightarrow a+b+c\le3\)
Vậy \(M\le3\)
Đẳng thức xảy ra khi a = b = c = 1
Ta có: \(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}=\frac{a^2+ab+1}{\sqrt{a^2+ab+2ab+c^2}}\ge\frac{a^2+ab+1}{\sqrt{a^2+ab+a^2+b^2+c^2}}=\sqrt{a^2+ab+1}\)
\(\sqrt{a^2+ab+1}=\sqrt{a^2+ab+a^2+b^2+c^2}=\sqrt{\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2+a^2+c^2}\)
\(=\frac{1}{\sqrt{5}}.\sqrt{\left(\frac{9}{4}+\frac{3}{4}+1+1\right)\left(\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2+a^2+c^2\right)}\)
\(\ge\frac{1}{\sqrt{5}}\sqrt{\left(\frac{3}{2}\left(a+\frac{b}{2}\right)+\frac{3}{2}b+a+c\right)^2}\)
\(=\frac{1}{\sqrt{5}}\left(\frac{5}{2}a+\frac{3}{2}b+c\right)\)
=> \(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}\ge\frac{1}{\sqrt{5}}\left(\frac{5}{2}a+\frac{3}{2}b+c\right)\)
Tương tự ta cũng chứng minh đc:
\(\frac{b^2+bc+1}{\sqrt{b^2+3bc+a^2}}\ge\frac{1}{\sqrt{5}}\left(\frac{5}{2}b+\frac{3}{2}c+a\right)\)
\(\frac{c^2+ca+1}{\sqrt{c^2+3ca+b^2}}\ge\frac{1}{\sqrt{5}}\left(\frac{5}{2}c+\frac{3}{2}a+b\right)\)
=> \(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}+\frac{b^2+bc+1}{\sqrt{b^2+3bc+a^2}}+\frac{c^2+ca+1}{\sqrt{c^3+3ca+b^2}}\ge\frac{1}{\sqrt{5}}\left(5a+5b+5c\right)\)
\(=\sqrt{5}\left(a+b+c\right)\)
Dấu "=" xảy ra <=> a = b = c =\(\frac{1}{\sqrt{3}}\)
Đặt \(P=\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ca}}+\frac{c}{\sqrt{c^2+abc}}\)
\(=\frac{a^2}{a\sqrt{a^2+8bc}}+\frac{b^2}{b\sqrt{b^2+8ca}}+\frac{c^2}{c\sqrt{c^2+abc}}\)
\(\ge\frac{\left(a+b+c\right)^2}{\left(a\sqrt{a^2+8bc}+b\sqrt{b^2+8ca}+c\sqrt{c^2+8ab}\right)}\)(Theo bất đẳng thức Bunhiacopxki dạng phân thức)
Ta có:
Suy ra 
Ta cần chứng minh \(a^3+b^3+c^3+24abc\le\left(a+b+c\right)^3\)
\(\Leftrightarrow a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\ge6abc\)
Đúng vì \(a^2b+b^2c+c^2a\ge3\sqrt[3]{a^3b^3c^3}=3abc\); \(ab^2+bc^2+ca^2\ge3\sqrt[3]{a^3b^3c^3}=3abc\)
Từ đó suy ra \(\left(a\sqrt{a^2+8bc}+b\sqrt{b^2+8ca}+c\sqrt{c^2+8ab}\right)\le\left(a+b+c\right)^2\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{\left(a\sqrt{a^2+8bc}+b\sqrt{b^2+8ca}+c\sqrt{c^2+8ab}\right)}\ge1\)
Vậy \(=\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ca}}+\frac{c}{\sqrt{c^2+abc}}\ge1\)
Đẳng thức xảy ra khi a = b = c
ta có:
\(\left(b-c\right)^2\ge0\Leftrightarrow b^2+4bc+4c^2\le3b^2+6c^2\Leftrightarrow\left(b+2c\right)^2\le3b^2+6c^2\)
\(\Leftrightarrow\frac{\left(b+2c\right)^2}{3b^2+6c^2}\le1\Leftrightarrow\frac{b+2c}{\sqrt{3b^2+6c^2}}\le1\Leftrightarrow\frac{a\left(b+2c\right)}{\sqrt{3b^2+6c^2}}\le a\)
cmtt =>\(\frac{a\left(b+2c\right)}{\sqrt{3b^2+6c^2}}+\frac{b\left(c+2a\right)}{\sqrt{3c^2+6a^2}}+\frac{c\left(a+2b\right)}{\sqrt{3a^2+6b^2}}\le a+b+c\left(Q.E.D\right)\)
dấu = xảy ra khi a=b=c
\(\sqrt{a^2+b^2+6c}=\sqrt{a^2+b^2+2c\left(a+b+c\right)}\)
\(=\sqrt{a^2+b^2+2c^2+2bc+2ca}=\sqrt{\left(a+c\right)^2+\left(b+c\right)^2}\)
\(\Rightarrow\frac{a+b}{\sqrt{\left(a+c\right)^2+\left(b+c\right)^2}}=\sqrt{\frac{\left(a+b\right)^2}{\left(a+c\right)^2+\left(b+c\right)^2}}\)
Đặt \(\left(\left(a+b\right)^2;\left(b+c\right)^2;\left(c+a\right)^2\right)=\left(x;y;z\right)\)
\(\Rightarrow P=\sum\sqrt{\frac{x}{y+z}}\)
Đến đây thì dễ rồi, bài toán cơ bản
\(\sqrt{x\left(y+z\right)}\le\frac{x+y+z}{2}\Rightarrow\frac{x\sqrt{y+z}}{\sqrt{x}}\le\frac{x+y+z}{2}\Rightarrow\sqrt{\frac{y+z}{x}}\le\frac{x+y+z}{2x}\)
\(\Rightarrow\sqrt{\frac{x}{y+z}}\ge\frac{2x}{x+y+z}\Rightarrow P\ge\sum\frac{2x}{x+y+z}=2\)
Dấu "=" ko xảy ra nên \(P>2\)
IMO, 2001
Đặt \(x=\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{c}{\sqrt{c^2+8ab}}\left(x;y;z\in\left(0;1\right)\right)\)
Để ý rằng \(\frac{a^2}{8bc}=\frac{x^2}{1-x^2};\frac{b^2}{8ac}=\frac{y^2}{1-y^2};\frac{c^2}{8ba}=\frac{z^2}{1-z^2}\)
=> \(\frac{1}{512}=\left(\frac{x^2}{1-x^2}\right)\left(\frac{y^2}{1-y^2}\right)\left(\frac{z^2}{1-z^2}\right)\)
Ta cần chứng minh \(x+y+z\ge1\)với \(x;y;z\in\left(0;1\right)\)và \(\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)=512\left(xyz\right)^2\left(1\right)\)
Giả sử ngược lại x+y+z<1
Theo BĐT AM-GM ta có:
\(\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)>\left[\left(x+y+z\right)^2-x^2\right]\left[\left(x+y+z\right)^2-y^2\right]\left[\left(x+y+z\right)^2-z^2\right]\)
\(=\left(x+x+y+z\right)\left(y+z\right)\left(x+y+z+y\right)\left(z+x\right)\left(z+z+x+y\right)\left(x+y\right)\)
\(\ge4\left(x^2yz\right)^{\frac{1}{4}}\cdot2\left(yz\right)^{\frac{1}{2}}\cdot4\left(y^2zx\right)^{\frac{1}{4}}\cdot2\left(xz\right)^{\frac{1}{2}}\cdot4\left(z^2xy\right)^{\frac{1}{4}}\cdot2\left(xy\right)^{\frac{1}{2}}=512\left(xyz\right)^2\)
Điều này mâu thuẫn với (1)
Vậy điều phản chứng là sai và ta có đpcm
Đấu đề bổ sung = 3 nhé
Xíu mk giải cho