Question

[monsiaur kite]

CM rằng:

\(\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab}\ge\frac{1}{5}\)

giúp ạ

Answer 1

[✎﹏ミ★꧁༺вєѕт↭ℓαυяιєℓ↭νи༻꧂★ミ.༻(Trưở...]

ta có:

\(\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2\ge0\) (luôn đúng)

\(\Leftrightarrow2a^4+2b^4+2c^4-2.\left(a^2b^2+b^2c^2+c^2a^2\right)\ge0\)

\(\Leftrightarrow a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\)

\(\Leftrightarrow a^4+b^4+c^4+2.\left(a^2b^2+b^2c^2+c^2a^2\right)\ge3.\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge3.\left(a^2b^2+b^2c^2+c^2a^2\right)\left(1\right)\)

+)ta có:\(a^2.b^2+b^2.c^2\ge2.\sqrt{a^2.b^2.b^2.c^2}=2ab^2c\)(BĐT cô si)

tương tự thì ta có:\(b^2c^2+c^2a^2\ge2abc^2;c^2a^2+a^2b^2\ge2a^2bc\)

\(\Rightarrow2.\left(a^2b^2+b^2c^2+c^2a^2\right)\ge2abc.\left(a+b+c\right)\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\left(a+b+c\right)\left(2\right)\)

từ 1 và 2,ta có: \(\left(a^2+b^2+c^2\right)^2\ge3.abc\left(a+b+c\right)\)

\(\Leftrightarrow1\ge3.abc\left(a+b+c\right)\)

\(\Leftrightarrow abc.\left(a+b+c\right)\le\frac{1}{3}\)

+)lại có:\(\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab}=\frac{a^4}{a^2+2a^2bc}+\frac{b^4}{b^2+2cab^2}+\frac{c^4}{c^2+2abc^2}\)

ÁP dụng BĐT cộng mẫu ta có:

\(\frac{a^4}{a^2+2a^2bc}+\frac{b^4}{b^2+2cab^2}+\frac{c^4}{c^2+2abc^2}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)+2abc\left(a+b+c\right)}\ge\frac{1}{1+2.\frac{1}{3}}=\frac{3}{5}\)

dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

\(\left(a^2+b^2+c^2\right)=1\)

\(\Leftrightarrow3a^2=1\)

\(\Leftrightarrow a=\pm\frac{1}{\sqrt{3}}\)

vậy \(\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab}\ge\frac{3}{5}\)dấu bằng xảy ra \(\Leftrightarrow a=b=c=\pm\frac{1}{\sqrt{3}}\)

Answer 2

[Xyz OLM]

Ta có \(P=\frac{a^4}{a^2+2a^2bc}+\frac{b^4}{b^2+2ab^2c}+\frac{c^4}{c^2+2abc^2}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2+2abc\left(a+b+c\right)}\)(BĐT Schwarz)

\(=\frac{1^2}{1+2abc\left(a+b+c\right)}\le\frac{1}{1+\frac{2\left(a+b+c\right)^3}{27}.\left(a+b+c\right)}=\frac{1}{1+\frac{2}{27}\left(a+b+c\right)^4}\)(BĐT Cauchy)

Lại có (a + b + c)² \(\le3\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow\left(a+b+c\right)^4\le9\left(a^2+b^2+c^2\right)^2=9\)

Khi đó P \(\ge\frac{1}{1+\frac{2}{27}.9}=\frac{1}{\frac{5}{3}}=\frac{3}{5}\left(\text{ĐPCM}\right)\)

Dấu "=" khi a = b = c = \(\frac{1}{\sqrt{3}}\)