Toán 10 bất đẳng thức

Mộc Nhãn said:

C1: Ta có: (b+c−a)2a2+(b+c)2=(1−2a)2a2+(1−a)2=4a2−4a+12a2−2a+1(b+c−a)2a2+(b+c)2=(1−2a)2a2+(1−a)2=4a2−4a+12a2−2a+1\frac{(b+c-a)^{2}}{a^{2}+(b+c)^{2}}=\frac{(1-2a)^2}{a^2+(1-a)^2}=\frac{4a^2-4a+1}{2a^2-2a+1}
Cần chứng minh 4a2−4a+12a2−2a+1≥23−54a25⇔2(3x−1)2(6x+1)≥0(luônđúng)4a2−4a+12a2−2a+1≥23−54a25⇔2(3x−1)2(6x+1)≥0(luônđúng)\frac{4a^2-4a+1}{2a^2-2a+1}\geq \frac{23-54a}{25}\Leftrightarrow 2(3x-1)^2(6x+1)\geq 0(luôn đúng)
Tương tự thì ta có: (b+c−a)2a2+(b+c)2+(c+a−b)2b2+(c+a)2+(a+b−c)2c2+(a+b)2≥23−52a25+23−52b25+23−52c25=35(đpcm)(b+c−a)2a2+(b+c)2+(c+a−b)2b2+(c+a)2+(a+b−c)2c2+(a+b)2≥23−52a25+23−52b25+23−52c25=35(đpcm)\frac{(b+c-a)^{2}}{a^{2}+(b+c)^{2}}+\frac{(c+a-b)^{2}}{b^{2}+(c+a)^{2}}+\frac{(a+b-c)^{2}}{c^{2}+(a+b)^{2}}\geq \frac{23-52a}{25}+\frac{23-52b}{25}+\frac{23-52c}{25}=\frac{3}{5}(đpcm)
C2: BĐT cần chứng minh ⇔1−(b+c−a)2a2+(b+c)2+1−(c+a−b)2b2+(c+a)2+1−(a+b−c)2c2+(a+b)2≤3−35⇔2ab+2aca2+(b+c)2+2bc+2abb2+(c+a)2+2ca+2cbc2+(a+b)2≤125⇔a(b+c)a2+(b+c)2+b(c+a)b2+(c+a)2+c(c+a)c2+(b+a)2≤65⇔a(1−a)2a2−2a+1+b(1−b)2b2−2b+1+c(1−c)2c2−2c+1≤65(1)⇔1−(b+c−a)2a2+(b+c)2+1−(c+a−b)2b2+(c+a)2+1−(a+b−c)2c2+(a+b)2≤3−35⇔2ab+2aca2+(b+c)2+2bc+2abb2+(c+a)2+2ca+2cbc2+(a+b)2≤125⇔a(b+c)a2+(b+c)2+b(c+a)b2+(c+a)2+c(c+a)c2+(b+a)2≤65⇔a(1−a)2a2−2a+1+b(1−b)2b2−2b+1+c(1−c)2c2−2c+1≤65(1)\Leftrightarrow 1-\frac{(b+c-a)^{2}}{a^{2}+(b+c)^{2}}+1-\frac{(c+a-b)^{2}}{b^{2}+(c+a)^{2}}+1-\frac{(a+b-c)^{2}}{c^{2}+(a+b)^{2}}\leq 3-\frac{3}{5}\Leftrightarrow \frac{2ab+2ac}{a^{2}+(b+c)^{2}}+\frac{2bc+2ab}{b^{2}+(c+a)^{2}}+\frac{2ca+2cb}{c^{2}+(a+b)^{2}}\leq \frac{12}{5}\Leftrightarrow \frac{a(b+c)}{a^2+(b+c)^2}+\frac{b(c+a)}{b^2+(c+a)^2}+\frac{c(c+a)}{c^2+(b+a)^2}\leq \frac{6}{5}\Leftrightarrow \frac{a(1-a)}{2a^2-2a+1}+\frac{b(1-b)}{2b^2-2b+1}+\frac{c(1-c)}{2c^2-2c+1}\leq \frac{6}{5}(1)
Ta thấy: 2c(1−c)≤(2c+1−c2)2=(1+c)24⇒2c2−2c+1=1−2c(1−c)≥1−(1+c)24=(1−c)(c+3)4⇒c(1−c)2c2−2c+1≤c(1−c)(1−c)(c+3)4=4cc+3=4.cc+3=4(1−3c+3)2c(1−c)≤(2c+1−c2)2=(1+c)24⇒2c2−2c+1=1−2c(1−c)≥1−(1+c)24=(1−c)(c+3)4⇒c(1−c)2c2−2c+1≤c(1−c)(1−c)(c+3)4=4cc+3=4.cc+3=4(1−3c+3)2c(1-c)\leq (\frac{2c+1-c}{2})^2=\frac{(1+c)^2}{4}\Rightarrow 2c^2-2c+1=1-2c(1-c)\geq 1-\frac{(1+c)^2}{4}=\frac{(1-c)(c+3)}{4}\Rightarrow \frac{c(1-c)}{2c^2-2c+1}\leq \frac{c(1-c)}{\frac{(1-c)(c+3)}{4}}=\frac{4c}{c+3}=4.\frac{c}{c+3}=4(1-\frac{3}{c+3})
Tương tự ta cần chứng minh: 4(1−3a+3)+4(1−3b+3)+4(1−3c+3)≤65⇔12(1a+3+1b+3+1c+3)≥12−65=545⇔1a+3+1b+3+1c+3≥9104(1−3a+3)+4(1−3b+3)+4(1−3c+3)≤65⇔12(1a+3+1b+3+1c+3)≥12−65=545⇔1a+3+1b+3+1c+3≥9104(1-\frac{3}{a+3})+4(1-\frac{3}{b+3})+4(1-\frac{3}{c+3})\leq \frac{6}{5}\Leftrightarrow 12(\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3})\geq 12-\frac{6}{5}=\frac{54}{5}\Leftrightarrow \frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}\geq \frac{9}{10}
Lại có: 1a+3+1b+3+1c+3≥9a+b+c+3=910⇒đpcm

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