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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)
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 \(\sqrt{x^2-xy+y^2}=\sqrt{\frac{1}{4}x^2+\frac{1}{2}xy+\frac{1}{4y^2}+\frac{3}{4}x^2-\frac{3}{2}xy+\frac{3}{4}y^2}\)
\(=\sqrt{\frac{1}{4}\left(x^2+2xy+y^2\right)+\frac{3}{4}\left(x^2-2xy+y^2\right)}=\sqrt{\frac{1}{4}\left(x+y\right)^2+\frac{3}{4}\left(x-y\right)^2}\)
\(P=\frac{y^3}{x^2+xy+y^2}+\frac{z^3}{y^2+zx+z^2}+\frac{x^3}{z^2+zx+x^2}\)
\(\Leftrightarrow P=\frac{y^4}{x^2y+xy^2+y^3}+\frac{z^4}{y^2z+z^2x+z^3}+\frac{x^4}{z^2x+zx^2+x^3}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^3+y^3+z^3+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}\)
\(\Leftrightarrow P\ge\frac{\left(x^2+y^2+z^2\right)^2}{\left(x+y+z\right)\left(x^2+y^2+z^2\right)}=\frac{x^2+y^2+z^2}{x+y+z}\ge3\)
Dấu "=" khi x=y=z=3
Ta sẽ chứng minh: \(\frac{a^3}{a^2+ab+b^2}\ge\frac{2a-b}{3}\) với a;b dương
Thật vậy, BĐT tương đương:
\(3a^3\ge\left(2a-b\right)\left(a^2+ab+b^2\right)\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng)
Áp dụng: \(\Rightarrow S\ge\frac{2x-y}{3}+\frac{2y-z}{3}+\frac{2z-x}{3}=\frac{x+y+z}{3}=3\)
\(S_{min}=3\) khi \(x=y=z=3\)
Đây là câu bđt của chuyên Quảng Nam vừa thi mà:vvv
Ta có: \(xy+yz+zx=xyz\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)\left(a,b,c>0\right)\)
Khi đó: \(H=\frac{a}{9b^2+1}+\frac{b}{9c^2+1}+\frac{c}{9a^2+1}\)
\(=\left(a+b+c\right)-\left(\frac{9ab^2}{9b^2+1}+\frac{9bc^2}{9c^2+1}+\frac{9ca^2}{9a^2+1}\right)\)
\(\ge1-\left(\frac{9ab^2}{6b}+\frac{9bc^2}{6c}+\frac{9ca^2}{6a}\right)\)
\(=1-\frac{3}{2}\left(ab+bc+ca\right)\ge1-\frac{3}{2}\cdot\frac{\left(a+b+c\right)^2}{3}=1-\frac{3}{2}\cdot\frac{1}{3}=\frac{1}{2}\)
Dấu "=" xảy ra khi: \(x=y=z=3\)
Vậy Min(H) = 1/2 khi x = y = z = 3
a/
Với mọi số thực x;y;z ta luôn có:
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+zx\)
\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2yz\ge3xy+3yz+3zx\)
\(\Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\) (đpcm)
b/
\(M=2\left(\frac{1}{x^2+y^2+z^2}+\frac{1}{xy+yz+zx}+\frac{1}{xy+yz+zx}\right)+\frac{1}{xy+yz+zx}\)
\(M\ge2.\frac{9}{x^2+y^2+z^2+xy+yz+zx+xy+yz+zx}+\frac{1}{\frac{\left(x+y+z\right)^2}{3}}\)
\(M\ge\frac{18}{\left(x+y+z\right)^2}+\frac{3}{\left(x+y+z\right)^2}=\frac{21}{\left(x+y+z\right)^2}=21\)
\(M_{min}=21\) khi \(x=y=z=\frac{1}{3}\)
Áp dụng bđt AM-GM ta có
\(P\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2.\left(yz+1\right)^2.\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=A\)
Ta có \(A=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Áp dụng bđt AM-GM ta có
\(A\ge3\sqrt[3]{8\sqrt{\frac{xyz}{xyz}}}=3.2=6\)
\(\Rightarrow P\ge6\)
Dấu "=" xảy ra khi x=y=z=\(\frac{1}{2}\)
Làm tiếp bài ღ๖ۣۜLinh's ๖ۣۜLinh'sღ] ★we are one★ chớ hình như bị ngược dấu ó.Do mình gà nên chỉ biết cô si mù mịt thôi ạ
\(3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
\(=3\sqrt[3]{\left(y+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}\right)\left(z+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}\right)\left(x+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}\right)}\)
\(\ge3\sqrt[3]{5\sqrt[5]{\frac{y}{256x^4}}\cdot5\sqrt[5]{\frac{z}{256y^4}}\cdot5\sqrt[5]{\frac{x}{256z^4}}}\)
\(=3\sqrt[3]{125\sqrt[5]{\frac{xyz}{256^3\left(xyz\right)^4}}}\)
\(=15\sqrt[3]{\sqrt[5]{\frac{1}{256^3\left(xyz\right)^3}}}\)
\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\left(\frac{x+y+z}{3}\right)^9}}\)
\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\frac{1}{2^9}}}=\frac{15}{2}\)
Dấu "=" xảy ra tại \(x=y=z=\frac{1}{2}\)
@Akai Haruma
@Nguyễn Việt Lâm
2 anh giúp e