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3, \(\sqrt{\frac{a}{b+c}}=\sqrt{\frac{a^2}{a\left(b+c\right)}}\Rightarrow\frac{1}{\sqrt{\frac{a}{b+c}}}=\sqrt{\frac{a\left(b+c\right)}{a^2}}.\)
Áp dụng bất đẳng thức Cô si ta có : \(\sqrt{\frac{a\left(b+c\right)}{a^2}}\le\frac{a+b+c}{2a}\Rightarrow\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\left(1\right).\)
Chứng minh tương tự ta có : \(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c}\left(2\right).\); \(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\left(3\right).\)
Cộng vế với vế các bất đẳng thức cùng chiều ta được:
\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2\left(a+b+c\right)}{a+b+c}=2.\)( đpcm )
dấu " = " xẩy ra khi a = b = c > 0
Câu 1 chuyên phan bội châu
câu c hà nội
câu g khoa học tự nhiên
câu b am-gm dựa vào hằng đẳng thử rồi đặt ẩn phụ
câu f đặt \(a=\frac{2m}{n+p};b=\frac{2n}{p+m};c=\frac{2p}{m+n}\)
Gà như mình mấy câu còn lại ko bt nha ! để bạn tth_pro full cho nhé !
Câu c quen thuộc, chém trước:
Ta có BĐT phụ: \(\frac{x^3}{x^3+\left(y+z\right)^3}\ge\frac{x^4}{\left(x^2+y^2+z^2\right)^2}\) \((\ast)\)
Hay là: \(\frac{1}{x^3+\left(y+z\right)^3}\ge\frac{x}{\left(x^2+y^2+z^2\right)^2}\)
Có: \(8(y^2+z^2) \Big[(x^2 +y^2 +z^2)^2 -x\left\{x^3 +(y+z)^3 \right\}\Big]\)
\(= \left( 4\,x{y}^{2}+4\,x{z}^{2}-{y}^{3}-3\,{y}^{2}z-3\,y{z}^{2}-{z}^{3 } \right) ^{2}+ \left( 7\,{y}^{4}+8\,{y}^{3}z+18\,{y}^{2}{z}^{2}+8\,{z }^{3}y+7\,{z}^{4} \right) \left( y-z \right) ^{2} \)
Từ đó BĐT \((\ast)\) là đúng. Do đó: \(\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\frac{x^2}{x^2+y^2+z^2}\)
\(\therefore VT=\sum\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\sum\frac{x^2}{x^2+y^2+z^2}=1\)
Done.
Chú ý đến giả thiết a + b + c = 1 ta viết được \(\frac{ab}{\sqrt{\left(1-c\right)^3\left(1+c\right)}}=\frac{ab}{\sqrt{\left(a+b\right)^2\left(1-c\right)\left(1+c\right)}}=\)\(\frac{ab}{\left(a+b\right)\sqrt{1-c^2}}=\frac{ab}{\left(a+b\right)\sqrt{\left(a+b+c\right)^2-c^2}}\)\(=\frac{ab}{\left(a+b\right)\sqrt{a^2+b^2+2\left(ab+bc+ca\right)}}\)
Mặt khác áp dụng bất đẳng thức Cauchy ta được \(a^2+b^2+2\left(ab+bc+ca\right)\ge2ab+2\left(ab+bc+ca\right)=\)\(2\left(ab+bc\right)+2\left(ab+ca\right)\)và \(a+b\ge2\sqrt{ab}\)
Từ đó dẫn đến \(\frac{ab}{\left(a+b\right)\sqrt{a^2+b^2+2\left(ab+bc+ca\right)}}\le\frac{ab}{2\sqrt{ab}\sqrt{2\left(ab+bc\right)+2\left(ab+ca\right)}}\)\(=\frac{1}{2}\sqrt{\frac{ab}{2\left(ab+bc\right)+2\left(ab+ca\right)}}\)
Mà theo bất đẳng thức quen thuộc \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\) ta có: \(\sqrt{\frac{ab}{2\left(ab+bc\right)+2\left(ab+ca\right)}}\le\sqrt{\frac{1}{4}\left(\frac{ab}{2\left(ab+bc\right)}+\frac{ab}{2\left(ab+ca\right)}\right)}\)
\(=\frac{1}{2\sqrt{2}}\sqrt{\frac{ab}{ab+bc}+\frac{ab}{ab+ca}}=\frac{1}{2\sqrt{2}}\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}\)
Từ đó ta có bất đẳng thức: \(\frac{ab}{\sqrt{\left(1-c\right)^3\left(1+c\right)}}\le\frac{1}{4\sqrt{2}}\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}\)(1)
Hoàn toàn tương tự, ta có: \(\frac{bc}{\sqrt{\left(1-a\right)^3\left(1+a\right)}}\le\frac{1}{4\sqrt{2}}\sqrt{\frac{b}{b+a}+\frac{c}{c+a}}\)(2) ; \(\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\le\frac{1}{4\sqrt{2}}\sqrt{\frac{c}{c+b}+\frac{a}{a+b}}\)(3)
Cộng theo vế 3 bất đẳng thức (1), (2), (3), ta được: \(\frac{ab}{\sqrt{\left(1-c\right)^3\left(1+c\right)}}+\frac{bc}{\sqrt{\left(1-a\right)^3\left(1+c\right)}}+\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\)\(\le\frac{1}{4\sqrt{2}}\left(\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}+\sqrt{\frac{b}{b+a}+\frac{c}{c+a}}+\sqrt{\frac{c}{c+b}+\frac{a}{a+b}}\right)\)
Ta cần chứng minh\(\frac{1}{4\sqrt{2}}\left(\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}+\sqrt{\frac{b}{b+a}+\frac{c}{c+a}}+\sqrt{\frac{c}{c+b}+\frac{a}{a+b}}\right)\le\frac{3\sqrt{2}}{8}\)
Hay \(\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}+\sqrt{\frac{b}{b+a}+\frac{c}{c+a}}+\sqrt{\frac{c}{c+b}+\frac{a}{a+b}}\le3\)
Áp dụng bất đẳng thức Bunhiacopxki ta được \(\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}+\sqrt{\frac{b}{b+a}+\frac{c}{c+a}}+\sqrt{\frac{c}{c+b}+\frac{a}{a+b}}\)
\(\le\sqrt{3\left(\frac{a}{a+c}+\frac{b}{b+c}+\frac{b}{b+a}+\frac{c}{c+a}+\frac{c}{c+b}+\frac{a}{a+b}\right)}=3\)
Vậy bất đẳng thức được chứng minh
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)
Sửa đề: \(\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\)
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gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)