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Đặt \(\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)\rightarrow\left(a;b;c\right)\)\(\Rightarrow\left\{{}\begin{matrix}a+b+c=1\\a;b;c>0\end{matrix}\right.\)
Và \(\dfrac{ab}{\sqrt{a^2+b^2+2c^2}}+\dfrac{bc}{\sqrt{b^2+c^2+2a^2}}+\dfrac{ca}{\sqrt{c^2+a^2+2b^2}}\le\dfrac{1}{2}\)
Ta có:\(\dfrac{ab}{\sqrt{a^2+b^2+2c^2}}=\dfrac{2ab}{\sqrt{\left(1+1+2\right)\left(a^2+b^2+2c^2\right)}}\)
\(\le\dfrac{2ab}{a+b+2c}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{ab+bc}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{bc+ac}{a+b}\right)\)
\(=\dfrac{1}{2}\left(a+b+c\right)=\dfrac{1}{2}\)
Dấu "=" khi \(a=b=c=\dfrac{1}{3}\Rightarrow x=y=z=\dfrac{1}{9}\)
Đặt \(\left(a,b,c\right)=\left(\sqrt{x},\sqrt{y},\sqrt{z}\right)\).
Xét 4 số m, n, p, q. Ta sẽ chứng minh \(\left(m+n+p+q\right)^2\le4\left(m^2+n^2+p^2+q^2\right)\) (*)
Thật vậy:
(*) \(\Leftrightarrow2\left(mn+np+pq+qm+mp+nq\right)\le3\left(m^2+n^2+p^2+q^2\right)\)
\(\Leftrightarrow\left(m-n\right)^2+\left(n-p\right)^2+\left(p-q\right)^2+\left(q-m\right)^2+\left(m-p\right)^2+\left(n-q\right)^2\ge0\) (luôn đúng).
Từ đó: \(\left(\sqrt{x}+\sqrt{y}+2\sqrt{z}\right)^2=\left(\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{z}\right)^2\le4\left(x+y+z+z\right)=4\left(x+y+2z\right)\)
\(\Leftrightarrow\sqrt{x}+\sqrt{y}+2\sqrt{z}\le2\sqrt{x+y+2z}\)
\(\Leftrightarrow\sqrt{\frac{xy}{x+y+2z}}=\frac{\sqrt{xy}}{\sqrt{x+y+2z}}\le\frac{2\sqrt{x}\sqrt{y}}{\sqrt{x}+\sqrt{y}+2\sqrt{z}}=\frac{2ab}{a+b+2c}\le\frac{1}{2}ab\frac{4}{\left(a+c\right)+\left(b+c\right)}\le\frac{1}{2}ab\left(\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)
Tương tự, ta có:
\(\sum\sqrt{\frac{xy}{x+y+2z}}\le\frac{1}{2}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=\frac{1}{2}\sum\left(\frac{ab}{a+c}+\frac{bc}{c+a}\right)=\frac{1}{2}\sum a=\frac{1}{2}\)
Bai 1: Ap dung BDT Bunhiacopxki ta co:
\(ax+by+cz+2\sqrt {(ab+ac+bc)(xy+yz+xz)} \)
\(≤ \sqrt {(a^2+b^2+c^2)(x^2+y^2+z^2)} + \sqrt {(ab+ac+bc)(xy+yz+zx)}+\sqrt {(ab+ac+bc)(xy+yz+zx)}\)
\(≤ \sqrt {(a^2+b^2+c^2+2ab+2ac+2bc)(x^2+y^2+z^2+2xy+2yz+2zx)}\)
\(= (a+b+c)(x+y+z)\)
=> \(Q.E.D\)
Tiep bai 4:Ta co:
BDT <=> \((2+y^2z)(2+z^2x)(2+x^2y)≥(2+x)(2+y)(2+z)\)
Sau khi khai trien con: \(2(z^2x+y^2z+x^2y)+x^2z+z^2y+y^2x≥xy+yz+zx+2x+2y+2z \)
Ap dung BDT Cosi ta co:
\(z^2x+x ≥ 2zx \) <=> \(z^2x≥2zx-x\)
Lam tuong tu ta co: \(2(z^2x+y^2z+x^2y)≥4xy+4yz+4zx-2x-2y-2z \)(1)
\(x^2z+{1\over z}≥2x \) <=> \(x^2z≥2x-xy \) (do xyz=1)
Lam tuong tu ta co: \(x^2z+z^2y+y^2x≥ 2y+2z+2x-xy-yz-zx\)(2)
Cong (1) voi (2) ta co: VT\(≥ 3(xy+yz+zx)\)(*)
Voi cach lam tuong tu ta cung duoc: VT\(≥ 3(x+y+z) \)(**)
Tu (*) va (**) suy ra : \(3 \)VT \(≥ 6(x+y+z)+3(xy+yz+zx) \)
<=> VT \(≥ 2(x+y+z)+xy+yz+zx\)
=> \(Q.E.D\)
Đặt \(\hept{\begin{cases}\sqrt{x}=p\\\sqrt{y}=q\\\sqrt{z}=r\end{cases}}\). Khi đó \(\hept{\begin{cases}p+q+r=1\\p,q,r>0\end{cases}}\)
và ta cần chứng minh \(\frac{pq}{\sqrt{p^2+q^2+2r^2}}+\frac{qr}{\sqrt{q^2+r^2+2p^2}}+\frac{rp}{\sqrt{r^2+p^2+2q^2}}\le\frac{1}{2}\)
Ta có: \(\frac{pq}{\sqrt{p^2+q^2+2r^2}}=\frac{2pq}{\sqrt{\left(1+1+2\right)\left(p^2+q^2+2r^2\right)}}\)
\(\le\frac{2pq}{p+q+2r}\le\frac{1}{2}\left(\frac{pq}{p+r}+\frac{pq}{q+r}\right)\)(Theo BĐT Cauchy-Schwarz và BĐT \(\frac{1}{u}+\frac{1}{v}\ge\frac{4}{u+v}\)) (1)
Hoàn toàn tương tự: \(\frac{qr}{\sqrt{q^2+r^2+2p^2}}\le\frac{1}{2}\left(\frac{qr}{q+p}+\frac{qr}{r+p}\right)\)(2); \(\frac{rp}{\sqrt{r^2+p^2+2q^2}}\le\frac{1}{2}\left(\frac{rp}{r+q}+\frac{rp}{p+q}\right)\)(3)
Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(\frac{pq}{\sqrt{p^2+q^2+2r^2}}+\frac{qr}{\sqrt{q^2+r^2+2p^2}}+\frac{rp}{\sqrt{r^2+p^2+2q^2}}\)\(\le\frac{1}{2}\left(\frac{r\left(p+q\right)}{p+q}+\frac{p\left(q+r\right)}{q+r}+\frac{q\left(r+p\right)}{r+p}\right)=\frac{1}{2}\left(p+q+r\right)=\frac{1}{2}\)(Do p + q + r = 1)
Đẳng thức xảy ra khi \(p=q=r=\frac{1}{3}\)hay \(x=y=z=\frac{1}{9}\)
\(VT=\sum\frac{x}{\sqrt{1+x^2}}=\sum\frac{x}{\sqrt{xy+xz+yz+x^2}}=\sum\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}\le\frac{1}{2}\sum\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)\(\Rightarrow VT\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+z}+\frac{y}{x+y}+\frac{z}{x+z}+\frac{z}{y+z}\right)=\frac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
Đặt \(\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)\rightarrow\left(a;b;c\right)\Rightarrow\hept{\begin{cases}a+b+c=1\\a;b;c>0\end{cases}}\)
Và \(\frac{ab}{\sqrt{a^2+b^2+2c^2}}+\frac{bc}{\sqrt{b^2+c^2+2a^2}}+\frac{ca}{\sqrt{c^2+a^2+2b^2}}\le\frac{1}{2}\)
Ta có :
\(\frac{ab}{a^2+b^2+2c^2}=\frac{2ab}{\sqrt{\left(1+1+2\right)\left(a^2+b^2+2c^2\right)}}\)
\(\le\frac{2ab}{a+b+2c}\le\frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)
Tương tự cho 2 BĐT còn lại roouf cộng theo vế :
\(VT\le\frac{1}{2}\left(\frac{ab+bc}{a+c}+\frac{ab+ac}{b+c}+\frac{bc+ac}{a+b}\right)=\frac{1}{2}\left(a+b+c\right)=\frac{1}{2}\)
Dấu " = " xảy ra khi \(a=b=c=\frac{1}{3}\Rightarrow x=y=z=\frac{1}{9}\)
Chúc bạn học tốt !!!