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Từ giả thiết \(ab+bc+ca=2abc\)suy ra \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\)thì \(\hept{\begin{cases}x+y+z=2\\x,y,z>0\end{cases}}\)và bất đẳng thức cần chứng minh trở thành \(\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^2}+\frac{z^3}{\left(2-z\right)^2}\ge\frac{1}{2}\)hay \(\frac{x^3}{\left(y+z\right)^2}+\frac{y^3}{\left(z+x\right)^2}+\frac{z^3}{\left(x+y\right)^2}\ge\frac{1}{2}\)
Áp dụng bất đẳng thức Bunyakovsky dạng phân thức ta được \(\frac{x^3}{\left(y+z\right)^2}+\frac{y^3}{\left(z+x\right)^2}+\frac{z^3}{\left(x+y\right)^2}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x\left(y+z\right)^2+y\left(z+x\right)^2+z\left(x+y\right)^2}\)\(=\frac{\left(x^2+y^2+z^2\right)^2}{x^2y+y^2x+x^2z+z^2x+y^2z+z^2y+6xyz}\)
Ta cần chứng minh\(\frac{\left(x^2+y^2+z^2\right)^2}{x^2y+y^2x+x^2z+z^2x+y^2z+z^2y+6xyz}\ge\frac{1}{2}\)\(\Leftrightarrow2\left(x^2+y^2+z^2\right)^2\ge x^2y+y^2x+x^2z+z^2x+y^2z+z^2y+6xyz\)
Thật vậy, theo một đánh giá quen thuộc ta có \(2\left(x^2+y^2+z^2\right)^2=2\left(x^2+y^2+z^2\right)\left(x^2+y^2+z^2\right)\)\(\ge\frac{2\left(x+y+z\right)^2\left(x^2+y^2+z^2\right)}{3}\)
Mà ta lại có \(\left(x+y+z\right)\left(x^2+y^2+z^2\right)=x^3+y^3+z^3+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y\)
Suy ra ta có \(\frac{2\left(x+y+z\right)^2\left(x^2+y^2+z^2\right)}{3}\ge\frac{4\left(x^3+y^3+z^3+x^2y+y^2x+x^2z+z^2x+y^2z+yz^2\right)}{3}\)
Ta cần chỉ ra được \(4\left(x^3+y^3+z^3+x^2y+y^2x+x^2z+z^2x+y^2z+yz^2\right)\)\(\ge3\left(x^2y+y^2x+x^2z+z^2x+y^2z+yz^2+6xyz\right)\)
Hay\(4\left(x^3+y^3+z^3\right)+x^2y+y^2x+x^2z+z^2x+y^2z+yz^2\ge18xyz\)
Áp dụng bất đẳng thức Cauchy ta được \(4\left(x^3+y^3+z^3\right)\ge12xyz\); \(x^2y+y^2z+z^2x\ge3xyz\); \(xy^2+yz^2+zx^2\ge3xyz\)
Cộng theo vế các bất đẳng thức trên ta được\(4\left(x^3+y^3+z^3\right)+x^2y+y^2x+x^2z+z^2x+y^2z+yz^2\ge18xyz\)
Vậy bất đẳng thức được chứng minh
Đẳng thức xảy ra khi \(a=b=c=\frac{3}{2}\)
Xí trước phần b
Ta có: \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
\(=\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\)
\(=\frac{bc}{a^2b+ca^2}+\frac{ca}{b^2c+ab^2}+\frac{ab}{c^2a+bc^2}\)
\(=\frac{b^2c^2}{a^2b^2c+a^2bc^2}+\frac{c^2a^2}{ab^2c^2+a^2b^2c}+\frac{a^2b^2}{a^2bc^2+ab^2c^2}\)
\(=\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{bc+ab}+\frac{\left(ab\right)^2}{ca+bc}\)
\(\ge\frac{\left(bc+ca+ab\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
Cách làm khác của phần b ngắn gọn hơn:)
Ta có; \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
\(=\frac{\frac{1}{a^2}}{a\left(b+c\right)}+\frac{\frac{1}{b^2}}{b\left(c+a\right)}+\frac{\frac{1}{c^2}}{c\left(a+b\right)}\)
\(=\frac{\left(\frac{1}{a}\right)^2}{ab+ca}+\frac{\left(\frac{1}{b}\right)^2}{bc+ab}+\frac{\left(\frac{1}{c}\right)^2}{ca+bc}\)
\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(\frac{ab+bc+ca}{abc}\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi: a = b = c = 1
Áp dụng BĐT Bu-nhi-a-cốp-ski, ta có:
\(\left(a+b+c\right)\left[\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right]\)
\(\ge\left(\frac{a}{ac+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\right)^2\) \(\left(1\right)\)
Lại có: \(\frac{a}{ac+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ac+a+abc}+\frac{b}{bc+b+1}+\frac{bc}{abc+bc+b}\) ( Do abc=1 )
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}\)
\(=1\) \(\left(2\right)\)
Từ (1) và (2) suy ra \(\left(a+b+c\right)\left[\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right]\ge1\)
Mà \(a;b;c>0\Rightarrow a+b+c>0\)
\(\Rightarrow\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\ge\frac{1}{a+b+c}\) (đpcm)
Chứng minh bất đẳng thức \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
Có: \(\left[\left(\frac{a}{\sqrt{x}}\right)^2+\left(\frac{b}{\sqrt{y}}\right)^2+\left(\frac{c}{\sqrt{z}}\right)^2\right]\left(\sqrt{x}^2+\sqrt{y}^2+\sqrt{z}^2\right)\ge\left(a+b+c\right)^2\) (Bunyakovsky)
\(\Leftrightarrow\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
abc = 1 => a^2.b^2.c^2 = 1
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}=\frac{a^2b^2c^2}{a^3\left(b+c\right)}+\frac{a^2b^2c^2}{b^3\left(c+a\right)}+\frac{a^2b^2c^2}{c^3\left(a+b\right)}\)
\(=\frac{\left(bc\right)^2}{ab+ac}+\frac{\left(ac\right)^2}{bc+ba}+\frac{\left(ab\right)^2}{ca+cb}\ge\frac{\left(ab+ac+bc\right)^2}{2\left(ab+ac+bc\right)}=\frac{\left(ab+ac+bc\right)}{2}\)
\(\ge\frac{3\sqrt[3]{ab.ac.bc}}{2}\)(Cauchy) \(=\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}a=b=c\\\frac{bc}{ab+ac}=\frac{ac}{bc+ba}+\frac{ab}{ca+cb}\Leftrightarrow\end{cases}a=b=c}\)
Mà abc=1 <=> a^3 = 1 <=> a=1 => b=c=a=1
https://diendantoanhoc.net/topic/80159-ch%E1%BB%A9ng-minh-frac1a2b3cfrac12a3bcfrac13bb2c-leqslant-frac316/
bạn tham khảo ở đây nhé
1)Áp dụng Bđt Am-Gm \(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}\cdot\frac{b}{a}}=2\)
2)Áp dụng Am-Gm \(a^2+b^2\ge2\sqrt{a^2b^2}=2ab;b^2+c^2\ge2bc;a^2+c^2\ge2ca\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
=>ĐPcm
3)(a+b+c)2\(\ge\)3(ab+bc+ca)
=>a2+b2+c2+2ab+2bc+2ca\(\ge\)3ab+3bc+3ca
=>a2+b2+c2-ab-bc-ca\(\ge\)0
=>2a2+2b2+2c2-2ab-2bc-2ca\(\ge\)0
=>(a2-2ab+b2)+(b2-2bc+c2)+(c2-2ac+a2)\(\ge\)0
=>(a-b)2+(b-c)2+(c-a)2\(\ge\)0
4)đề đúng \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)
Đặt ⎧⎪⎨⎪⎩a+b−c=xb+c−a=yc+a−b=z(x,y,z>0){a+b−c=xb+c−a=yc+a−b=z(x,y,z>0)
⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩a=z+x2b=x+y2c=y+z2⇒{a=z+x2b=x+y2c=y+z2
⇒√a(1b+c−a−1√bc)=√2(z+x)2(1y−2√(x+y)(y+z))≥√x+√z2(1y−2√xy+√yz)=√x+√z2y−1√y⇒a(1b+c−a−1bc)=2(z+x)2(1y−2(x+y)(y+z))≥x+z2(1y−2xy+yz)=x+z2y−1y
Tương tự
⇒∑√a(1b+c−a−1√bc)≥∑√x+√z2y−∑1√y⇒∑a(1b+c−a−1bc)≥∑x+z2y−∑1y
⇒VT≥∑[x√x(y+z)]2xyz−∑√xy√xyz≥2√xyz(x+y+z)2xyz−x+y+z√xyz≐x+y+z√xyz−x+y+z√xyz=0⇒VT≥∑[xx(y+z)]2xyz−∑xyxyz≥2xyz(x+y+z)2xyz−x+y+zxyz≐x+y+zxyz−x+y+zxyz=0
(∑√xy≤x+y+z,x√x(y+z)≥2x√xyz)(∑xy≤x+y+z,xx(y+z)≥2xxyz)
dấu = ⇔x=y=z⇔a=b=c
\(P=\frac{a^2}{b^2+2bc}+\frac{b^2}{c^2+2ac}+\frac{c^2}{a^2+2ab}\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\)
Dấu "=" xảy ra khi \(a=b=c\)