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ta có \(\sqrt{x^2-xy+y^2}=\sqrt{\frac{1}{4}\left(x+y\right)^2+\frac{3}{4}\left(x-y\right)^2}\ge\sqrt{\frac{1}{4}\left(x+y\right)^2}=\frac{1}{2}\left(x+y\right)\)
tương tự ta có các trường hợp còn lại và ta có
\(S\ge\frac{1}{2}\left(\frac{x+y}{x+y+2z}+\frac{y+z}{y+z+2x}+\frac{z+x}{z+x+2y}\right)\)
đặt \(x+y=a;y+z=b;z+x=c\)
=> \(S\ge\frac{1}{2}\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\)
đặt \(A=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ca+ca}\)
Áp dụng bđt svác sơ ta có
\(A\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
mạt khác Áp dụng bđt cô si ta có
\(\hept{\begin{cases}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ac\end{cases}}\)
=> \(a^2+b^2+c^2\ge2\left(ab+bc+ca\right)\)
=> \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
=> \(A\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)
=> \(S\ge\frac{3}{4}\)
dấu = xảy ra <=> x=y=z>o
ta có \(\sqrt{x^2-xy+y^2}=\sqrt{\frac{1}{4}x^2+\frac{1}{2}xy+\frac{1}{4y^2}+\frac{3}{4}x^2-\frac{3}{2}xy+\frac{3}{4}y^2}\)
\(=\sqrt{\frac{1}{4}\left(x^2+2xy+y^2\right)+\frac{3}{4}\left(x^2-2xy+y^2\right)}=\sqrt{\frac{1}{4}\left(x+y\right)^2+\frac{3}{4}\left(x-y\right)^2}\)
Ta có : 2P = \(\frac{\sqrt{4x^2-4xy+4y^2}}{x+y+2z}+\frac{\sqrt{4y^2-4yz+4z^2}}{y+z+2x}+\frac{\sqrt{4z^2-4zx+4x^2}}{z+x+2y}\)
\(=\frac{\sqrt{\left(2x-y\right)^2+\left(\sqrt{3}y\right)^2}}{x+y+2z}+\frac{\sqrt{\left(2y-z\right)^2+\left(\sqrt{3}z\right)^2}}{y+z+2x}+\frac{\sqrt{\left(2z-x\right)^2+\left(\sqrt{3}x\right)^2}}{z+x+2y}\)
Lại có \(\frac{\sqrt{\left[\left(2x-y\right)^2+\left(\sqrt{3}y\right)^2\right]\left[\left(1^2+\left(\sqrt{3}\right)^2\right)\right]}}{x+y+2z}\ge\frac{\left[\left(2x-y\right).1+3y\right]}{x+y+2z}=\frac{2\left(x+y\right)}{x+y+2z}\)
=> \(\sqrt{\frac{\left(2x-y\right)^2+\left(\sqrt{3}y\right)^2}{x+y+2z}}\ge\frac{x+y}{x+y+2z}\)(BĐT Bunyakovsky)
Tương tự ta đươc \(2P\ge\frac{x+y}{x+y+2z}+\frac{y+z}{2x+y+z}+\frac{z+x}{2y+z+x}\)
Đặt x + y = a ; y + z = b ; x + z = c
Khi đó \(2P\ge\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)
\(\ge\left(a+b+c\right).\frac{9}{2\left(a+b+c\right)}-3\ge\frac{9}{2}-3=\frac{3}{2}\)
=> \(P\ge\frac{3}{4}\)
Dấu "=" xảy ra <=> x = y = z
bài 8 : bỏ dấu hoặc rồi tính
a;( 17 - 299) + ( 17 - 25 + 299)
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:
\(\sum\sqrt{\frac{xy}{x+y+2z}}\le\frac{1}{2}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=\frac{1}{2}\sum\left(\frac{ab}{a+c}+\frac{bc}{c+a}\right)=\frac{1}{2}\sum a=\frac{1}{2}\)
gọi P là cái 1/x+1/y+1/z nha
1) (1/x+1/y+1/z)^2 = 1/x^2 + 1/y^2 + 1/z^2 + 2/(xy) + 2/(yz) + 2/(zx)
---> 3 = P + 2(x+y+z)/(xyz) = P + 2 ---> P = 1
Ta có \(\frac{\sqrt{x^2+2y^2}}{xy}=\sqrt{\frac{1}{y^2}+\frac{2}{x^2}}\)
Áp dụng BĐT Buniacoxki ta có
\(\sqrt{\left(\frac{1}{y^2}+\frac{2}{x^2}\right)\left(1+2\right)}\ge\sqrt{\left(\frac{1}{y}+\frac{2}{x}\right)^2}=\frac{1}{y}+\frac{2}{x}\)
=> \(\sqrt{3}A\ge3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3\)
=> \(A\ge\sqrt{3}\)
\(MinA=\sqrt{3}\)khi x=y=z=3
Thấy cái đề mà thấy khiếp ...
Ta có : \(x^2-xy+y^2=\frac{3}{4}\left(x^2-2xy+y^2\right)+\frac{1}{4}\left(x^2+2xy+y^2\right)\)
\(=\frac{3}{4}\left(x-y\right)^2+\frac{1}{4}\left(x+y\right)^2\ge\frac{1}{4}\left(x+y\right)^2\)
\(\Rightarrow\sqrt{x^2-xy+y^2}\ge\frac{x+y}{2}\)
Tương tự \(\sqrt{y^2-yz+z^2}\ge\frac{y+z}{2}\)
\(\sqrt{z^2-zx+x^2}\ge\frac{x+z}{2}\)
Do đó : \(2S\ge\frac{x+y}{x+y+2z}+\frac{y+z}{y+z+2x}+\frac{x+z}{x+z+2y}\)
\(\Rightarrow2S+3\ge\left(1+\frac{x+y}{x+y+2z}\right)+\left(1+\frac{y+z}{y+z+2x}\right)+\left(1+\frac{x+z}{x+z+2y}\right)\)
\(=2\left(x+y+z\right)\left(\frac{1}{x+y+2z}+\frac{1}{y+z+2x}+\frac{1}{x+z+2y}\right)\)
\(\ge2\left(x+y+z\right).\frac{9}{4\left(x+y+z\right)}\)\(=\frac{9}{2}\)
(Áp dụng bđt Cô-si dạng engel cho 3 số)
\(\Rightarrow2S+3\ge\frac{9}{2}\)
\(\Rightarrow S\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)
Vậy ..............